NCERT Exemplar Class 11 Physics Solutions Chapter 5 Laws of Motion 2021

NCERT Exemplar Class 11 Physics Solutions Chapter 5 Laws of Motion

Edited By Safeer PP | Updated on Aug 08, 2022 04:03 PM IST

NCERT Exemplar Class 11 Physics solutions chapter 5 derives reference from the previous chapters of motion to describe the motion of a particle in space quantitatively. This chapter answers the pressing question- What governs the motion of bodies? NCERT Exemplar Class 11 Physics chapter 5 Solutions talks about the external agencies of force to keep a body in motion or stop an object's motion. The possibility that this external force may or may not be in contact with the object is also one aspect of motion. This chapter deals with the concept of inertia and the building blocks of the world's working- The laws of motion. The given chapter consists of an in-depth description of these laws, which govern everything around us.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy

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NCERT Exemplar Class 11 Physics Solutions Chapter 5 MCQI
NCERT Exemplar Class 11 Physics Solutions Chapter 5 MCQII
NCERT Exemplar Class 11 Physics Solutions Chapter 5 Very Short Answer
NCERT Exemplar Class 11 Physics Solutions Chapter 5 Short Answer
NCERT Exemplar Class 11 Physics Solutions Chapter 5 Long Answer
What Will The Students Learn in NCERT Exemplar Class 11 Physics Solutions Chapter 5?
NCERT Exemplar Class 11 Solutions
NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
Important Topics To Cover For Exams in NCERT Exemplar Solutions For Class 11 Physics Chapter 5 Laws of Motion

NCERT Exemplar Class 11 Physics Solutions Chapter 5 MCQI

Question:5.1

A ball is travelling with uniform translatory motion. This means that
(a) it is at rest.
(b) the path can be a straight line or circular and the ball travels with uniform speed.
(c) all parts of the ball have the same velocity (magnitude and direction) and the velocity is constant.
(d) the centre of the ball moves with constant velocity and the ball spins about its centre uniformly.

Answer:

The answer is the option (c) All particles of the ball have the same velocity (magnitude and direction), and the velocity is constant.
Explanation: Uniform motion or uniform translatory motion is the one in which all the particles of a body move in a straight line and with the same velocity.

Question:5.2

A metre scale is moving with uniform velocity. This implies
(a) the force acting on the scale is zero, but a torque about the centre of mass can act on the scale.
(b) the force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero.
(c) the total force acting on it need not be zero but the torque on it is zero.
(d) neither the force nor the torque need to be zero.

Answer:

The answer is the option (b) The net force acting on the scale is zero, and the net torque acting about the centre of mass of the scale is also zero.
Explanation: There is no change in velocity as the meter scale is moving with uniform velocity
Therefore, by Newton’s second law, its acceleration will be zero.
$\overrightarrow{F } _{net} = m \times 0 = 0$
Hence, the net or resultant force acting on the body must be 0.
Now, $\overrightarrow{\tau } =\overrightarrow{ r } \times \overrightarrow{ F}$
Therefore, toque must be zero as well.
Hence it is obvious that for an object to move with uniform velocity, both net force and torque must be zero.

Question:5.3

A cricket ball of mass 150 g has an initial velocity $u = (3 \widehat{i} + 4 \widehat{j} ) m s ^{-1}$
and a final velocity $v= -(3 \widehat{i} + 4 \widehat{j} ) m s ^{-1}$ after being hit. The change
in momentum (final momentum-initial momentum) is (in kg m s¹ )
(a) zero
(b) $-(0.45 \widehat{i} + 0.6 \widehat{j} )$
(c) $-(0.9 \widehat{i} + 1.2 \widehat{j} )$
(d) $-5( \widehat{i} + \widehat{j} )$

Answer:

The answer is the option (c) $-(0.9 \widehat{i} + 1.2 \widehat{j} )$
Explanation: Here, m = 150g = 0.15 kg
$u = (3 \widehat{i} + 4 \widehat{j} ) m s ^{-1}$ &
$v= -(3 \widehat{i} + 4 \widehat{j} ) m s ^{-1}$
Now, we know that
Change in momentum $\Delta (\overrightarrow{p})$ = Final momentum – initial momentum
$=m\overrightarrow{v}-m\overrightarrow{u}$
$=m(\overrightarrow{v}-\overrightarrow{u})$
$= 0.15 [ -(3\widehat{i} + 4\widehat{j}) -(3\widehat{i} + 4\widehat{j})]$
$= 0.15 [ -6\widehat{i} - 8\widehat{j}) ]$
Therefore, $\Delta \overrightarrow{p} = 0.9\widehat{i} -1.2\widehat{j} = - [0.9\widehat{i} + 1.2\widehat{j}]$

Question:5.4

In the previous problem (5.3), the magnitude of the momentum transferred during the hit is
(a) Zero
(b) 0.75 $kg m s^{-1}$
(c) 1.5 $kg m s^{-1}$
(d) 14 $kg m s^{-1}$

Answer:

The answer is the option (c) 1.5 $kg m s^{-1}$
Explanation: Change in momentum $(\Delta \overrightarrow{p})$ = Final momentum – initial momentum
$(\Delta \overrightarrow{p})=-[ 0.9\widehat{i}+1.2\widehat{j} ]$ Thus, its magnitude will be,
$\left |\Delta \overrightarrow{p} \right |=\sqrt{(0.9)^{2}+(1.2)^{2}}$
$\left |\Delta \overrightarrow{p} \right |=\sqrt{0.81+1.44}$
$\left |\Delta \overrightarrow{p} \right |=\sqrt{2.25}=1.5kgms^{-1}$

Question:5.5

Conservation of momentum in a collision between particles can be understood from
(a) conservation of energy.
(b) Newton’s first law only.
(c) Newton’s second law only.
(d) both Newton’s second and third law.

Answer:

. The answer is the option (d) Both Newton’s second and third law
Explanation: (ii) $\frac{\overrightarrow{dp}}{dt}=\overrightarrow{F}_{ext}$ ……. (By Newton’s second law)

By the law of conservation of momentum
$\overrightarrow{F}_{ext} =0$
Thus, $\frac{\overrightarrow{dp}}{dt}=0$ hence, $\overrightarrow{p}$ = constant

(ii) Newton’s third law states that every action has an equal, opposite and instantaneous reaction, i.e., action force is equal to reaction force in magnitude but opposite in direction.
$\overrightarrow{F}_{12}=-\overrightarrow{F}_{21}\left ( \overrightarrow{F}_{ext}=0 \right )$
$\frac{\overrightarrow{dp}_{12}}{dt}=-\frac{\overrightarrow{dp}_{21}}{dt}or \overrightarrow{dp}_{12}=-\overrightarrow{dp}_{21}$
$\overrightarrow{dp}_{12}+\overrightarrow{dp}_{21}=0$
Hence, option (d).

Question:5.6

A hockey player is moving northward and suddenly turns westward with the same speed to avoid an opponent. The force that acts on the player is
(a) frictional force along westward.
(b) muscle force along southward.
(c) frictional force along south-west.
(d) muscle force along south-west.

Answer:

The answer is the option (c) friction force along the south-west.
Explanation: The force on the player is a result of the rate of change of momentum. The direction of change in momentum, i.e., $\overrightarrow{p_{2}}-\overrightarrow{p_{1}}$ is the same as the direction of force acting on the player and is towards the south-west, viz., the direction of the force on the player.

Question:5.7

A body of mass 2kg travels according to the law $x(t ) = pt + qt^{2} + rt^{3}$ where $p = 3 ms^{-1}, q = 4 ms^{-2}$ and $r=5ms^{-3}$ . The force acting on the body at t = 2 seconds is
(a) 136 N
(b) 134 N
(c) 158 N
(d) 68 N

Answer:

The answer is the option (a) 136N
Explanation:
$x(t ) = pt + qt^{2} + rt^{3}$
$p = 3 ms^{-1}, q = 4 ms^{-2}$ $r=5ms^{-3}$
$x(t) = 3t + 4t^{2} + 5t^{3}$
thus, $\frac{dx(t)}{dt } = 3 + 8t + 15t^{2}$
$\frac{d^{2}x(t)}{dt^{2}}=0+8+30t$
$\left [\frac{d^{2}x(t)}{dt^{2}} \right ]_{t=2}=8+30\times 2=68ms^{-2}$
Now, we know that,
$\overrightarrow{F}=m\overrightarrow{a}=m.\frac{d^{2}x}{dt^{2}}$
Thus,
$\overrightarrow{F}=2 \times 68$
$=136\ N$

Question:5.8

A body of mass 5kg is acted upon by a force $F= (-3\widehat{i}+4\widehat{j})N$ . If its intial velocity at t=0 is $v= (6\widehat{i}-12\widehat{j})ms^{-1}$ , the time at which it will just have velocity along the y-axis is
(a) never
(b) 10 s
(c) 2 s
(d) 15 s

Answer:

The answer is the option (b) 10s
Explanation:
Initial velocity $(u) = (6\widehat{i} + 12\widehat{j}) ms^{-1}$
Force, $F= (-3\widehat{i}+4\widehat{j})N$ , Mass (m) = 5kg
$\overrightarrow{a} = \frac{\overrightarrow{F}}{m} = (\frac{-3}{5}\widehat{i}+ \frac{4}{5}\widehat{j}) ms^{-2}$
Since there is only Y component in final velocity and the X component is zero,
$v_{x} = u_{x} + a_{x}t$
$0 = 6 + \frac{-3}{5} t \rightarrow \frac{3}{5}t = 6$

t = 10s.

Question:5.9

A car of mass m starts from rest and acquires a velocity along east $v=v\widehat{i}$ in two seconds. Assuming the car moves with uniform acceleration, the force exerted on the car is
a) $\frac{mv}{2}$ eastward and is exerted by the car engine
b) $\frac{mv}{2}$ eastward and is due to the friction on the tyres exerted by the road
c) more than $\frac{mv}{2}$ eastward exerted due to the engine and overcomes the friction of the road
d) $\frac{mv}{2}$ exerted by the engine

Answer:

The answer is the option (b) $\frac{mv}{2}$ eastward and is due to friction on tyres exerted by the road.
Explanation: Here, $u = 0, \overrightarrow{v} = vi, t = 2, m = m$
We know that, v = u + at
$\overrightarrow{v }\widehat{i}= 0 + \overrightarrow{a} \times 2$
$\overrightarrow{a} =\frac{\overrightarrow{v }\widehat{i}}{2}$
$\overrightarrow{F} = m\overrightarrow{a} = \frac{m\overrightarrow{v}}{2}\widehat{i}$
The force by the engine is an internal force.

NCERT Exemplar Class 11 Physics Solutions Chapter 5 MCQII

Question:5.10

The motion of a particle of mass m is given by x = 0 for t < 0 s, x( t ) = A sin 4p t for 0 < t <(1/4) s (A > o), and x = 0 for t >(1/4) s. Which of the following statements is true?
(a) The force at t = (1/8) s on the particle is $-16\pi ^2 A m.$
(b) The particle is acted upon by on impulse of magnitude $4\pi ^2 A m.$ at t = 0 s and t = (1/4) s.
(c) The particle is not acted upon by any force.
(d) The particle is not acted upon by a constant force.
(e) There is no impulse acting on the particle.

Answer:

The correct answer is: -
(a) The force at $t =\frac{1}{8}$ s on the particle is $-16\pi ^2 A m.$
(b) The particle is acted upon by on impulse of magnitude $4 \pi ^2 A m.$ at t = 0s and $\left (t =\frac{1}{4} \right )s$
(d) The particle is not acted upon by a constant force.
Explanation: Here, mass = m
Thus, x(t) = 0, for t<0
$x(t) = A \sin4\pi t, for 0 < t < \left ( \frac{1}{4} \right ) s$
and
$x(t) =0, for t > \left ( \frac{1}{4} \right ) s$
Now,
For
$0< t < \left ( \frac{1}{4} \right ) s, x(t)=A\sin 4\pi t$
$v =\frac{ dx}{dt }= 4A\pi \cos4\pi t$
$a =\frac{ dv}{dt}= -16 \pi ^2A \sin 4\pi t$
$F(t)=ma(t)= -16 \pi ^2A m\sin 4\pi t$
Now, it’s clear that force is a function of time, hence opt (d) is verified.
(a) At $t=\left (\frac{1}{8} \right )s$
$a = -16\pi A\sin \frac{\pi}{2}$
$a(t) = -16\pi ^{2}A$
thus, force at $t =\frac{1}{8}$
$F = ma = m(-16\pi ^2A)$
Hence, $F = -16\pi ^{2}AmN.$
Hence, opt (a)
(b) Impulse = Change in momentum between the values t = 0s and $\left (t =\frac{1}{4} \right )s$
Now we already know from above that, F(t) varies from 0 at t = 0s to the max. value at $F (t)= -16\pi ^{2}AmN.$ at $\left (t =\frac{1}{4} \right )s$ & by the eq. $\overrightarrow{I}=\overrightarrow{F}t$ at $\left (t =\frac{1}{4} \right )s$ .
$|I|=16\pi^2Am\times\frac{1}{4}=4\pi^2Am$
Hence opt (b).

Question:5.11

In Figure, the co-efficient of friction between the floor and the body B is 0.1. The co-efficient of friction between the bodies B and A is 0.2. A force F is applied as shown on B. The mass of A is $\frac{m}{2}$ and of B is m. Which of the following statements are true?
capture-23
(a) The bodies will move together if F = 0.25 mg.
(b) The body A will slip with respect to B if F = 0.5 mg.
(c) The bodies will move together if F = 0.5 mg.
(d) The bodies will be at rest if F = 0.1 mg.
(e) The maximum value of F for which the two bodies will move together is 0.45 mg.

Answer:

The correct answer is: -
(a) The bodies will move together at if $\overrightarrow{ F} = 0.25 mg$ .
(b) The body A will slip with respect to B if $\overrightarrow{ F} = 0.5 mg$ .
(d)The bodies will be at rest if $\overrightarrow{ F} = 0.1 mg$ .
(e) The maximum value of $\overrightarrow{ F}$ for which the two bodies will move together is 0.45 mg.
Explanation: By opt (e) the max. force by which bodies move together is 0.45 mg Newton.
mA = $\frac{m}{2}$ = mB = m
Consider the acceleration of the bodies A and B to be ‘a’.
Both the bodies A & B will keep moving by force F until the force of friction between their surface is larger or equal to 0 than the force acting on A.
$a=\frac{F-f_{1}}{m_{A}-m_{B}}=\frac{F-f_{1}}{\frac{m}{2}+m}=\frac{2(F-f_{1})}{3m}$
capture-24

Thus, the force on
$A=m_{A}a=\frac{m}{2}\frac{\left ( F-f_{1} \right )}{3m}$
Hence, the force on A is
$F_{AB}=\frac{\left ( F-f_{1} \right )}{3}$
The body A will move along with body B only if F_AB is equal or smaller than f₂.
Hence, $F_{AB}=f_{2}$
$\mu N =\frac{\left ( F-f_{1} \right )}{3}$
$0.2\times m_{A}g =\frac{\left ( F-f_{1} \right )}{3}$ ……… (i)

N is the reaction force by B on A
$f_{1} = \mu NB = \mu (m_{A} + m_{B}) g$ ……… (here, N_B is the normal reactn on B along with A by the surface)
$0.1 \times (m_{A} + m_{B}) g$
$f_{1} = 0.1 \times \frac{3}{2} mg = 0.15mg$ ………… (ii)
Now,
$F - f_{1} = 3 \times 0.2 mAg$ ……….. [from (i)]
$F - 0.15mg = 0.6 \times \frac{m}{2} g$
$F_{max} = 0.3mg + 0.15mg = 0.45mg$ …………. (iii)
Thus, the max force on B is 0.45mg, therefore, A & B can move together.
Hence, opt (e).
Both the bodies A & B will keep moving by force F until the force of friction between their surface is larger or equal to 0 than the force acting on A, i.e., 0.45 mg Newton, hereby opt (c) is rejected.
Now, for opt (d), the minimum force which can move A & B together is,
$F_{in} \geq f_{1}+f_{2}$
$\geq 0.15 mg+0.2\times \frac{m}{2}g$ ….. [from (i) & (ii)]
$F_{in}\geq 0.25 mg$ Newton
Since 0.1mg<0.25 mg, opt (d) is verified which states that the body will be at rest if $\overrightarrow{F} = 0.1 mg.$

Question:5.12

Mass $m_{1}$ moves on a slope making an angle $\theta$ with the horizontal and is attached to mass $m_{2}$ by a string passing over a frictionless pulley as shown in Figure. The co-efficient of friction between $m_{1}$ and the sloping surface is $\mu$ .
capture-31
Which of the following statements are true?
(a) If $m_{2} > m_{1} \sin \theta$ , the body will move up the plane.
(b) If $m_{2} > m_{1} \sin \theta +\mu \cos \theta$ , the body will move up the plane.
(c) If $m_{2} < m_{1} \sin \theta +\mu \cos \theta$ , the body will move up the plane.
(d) If $m_{2} < m_{1} \sin \theta -\mu \cos \theta$ , the body will move down the plane.

Answer:

The correct answer is: -
(b) If $m_{2} > m_{1} \sin \theta +\mu \cos \theta$ , the body will move up the plane.
(d) If $m_{2} < m_{1} \sin \theta -\mu \cos \theta$ , the body will move down the plane.
Explanation:

Case I-
Considering the first case, normal reaction (N) = $m_{1}g\cos \theta$
Now, $f = \mu N = \mu m_{1}g\cos \theta$ , which becomes,
$T - m_{1}g\ sin \theta - \mu m_{1}g\cos \theta$
$= m_{1}a$
m₁ will be up, and m₂ will be down when $m_{2}g - (m_{1}g\cos \theta + f) > 0$
$m_{2}g - m_{1}g \sin \theta - g \mu m_{1}g \cos g \theta > 0$
$m_{2}g > m_{1}g (\sin\theta + \mu \cos \theta )$
or $m_{2} > m_{1} (\sin\theta + \mu \cos \theta )$
hereby, opt (b) is verified and opt (a) is rejected.
Case II-
Now, considering that if the body m1 moves down and the body m2 up then, direct of f becomes upward (i.e., opposite to the motion).
$-f + m_{1}g\sin\theta > m_{2}g$
$-\mu m _{1}g \cos \theta + m_{2}g \sin \theta > m _{2}g$
$m_{1} (-\mu \cos \theta + \sin\theta ) > m_2$
$m_{2} < m_{1} (\sin \theta - \mu \cos \theta )$
Hence, option (d) is verified here, as well as opt (c) is rejected.

Question:5.13

In Figure, a body A of mass m slides on plane inclined at angle $\theta _{1}$ to the horizontal and $\mu _{1}$ is the coefficient of friction between A and the plane. A is connected by a light string passing over a frictionless pulley to another body B, also of mass m, sliding on a frictionless plane inclined at angle $\theta _{2}$ to the horizontal. Which of the following statements are true?
capture-33
(a) A will never move up the plane.
(b) A will just start moving up the plane when $\mu = \frac{\sin \theta _{2} - \sin \theta _{1}}{\cos \theta _{1}}$
(c) For A to move up the plane, $\theta _{2}$ must always be greater than $\theta _{1}$ .
(d) B will always slide down with constant speed.

Answer:

The correct answer is the option:
(b) A will just start moving up the plane when $\mu = \frac{\sin \theta _{2} - \sin \theta _{1}}{\cos \theta _{1}}$
(c) For A to move up the plane, $\theta _{2}$ must always be greater than $\theta _{1}$ .
Explanation:

In the fig. below, $\mu$ = coefficient of friction of A whereas B is on a frictionless surface.
(i) when A is about to start
$f =\mu N_{1} = \mu mg \cos \theta _{1}$
$mg sin\theta _{1} +f=mg \sin \theta _{2}$
When A just start moving upward
$mg sin\theta _{1} +\mu mg cos\theta _{1}=mg \sin \theta _{2}$
$\mu=\frac{sin\theta_2-sin\theta_1}{cos\theta_1}$
If on the plane, the body A moves upward, and B moves downward
$mg \sin \theta _{2} - mg \sin \theta _{1 }>0$
$\sin \theta _{2} - \sin \theta _{1 }>0$
$\theta _{2} - \theta _{1 }>0$
Hence, opt (c) is verified and opt (a) is rejected.

(ii) If the body B moves upward and A moves downward then,
$mg\sin \theta_{ 2} -f - mg\sin \theta_{ 2} > 0$
$mg\sin \theta _{1} - \mu mg \cos \theta_{1} - mg\sin \theta _{2} > 0$
$\sin \theta _{1} - \sin \theta_{1} > \mu \cos\theta_{ 1}$
Since $\theta _{1}$ and $\theta _{2}$ are acute angles, $\theta _{1}$ > $\theta _{2}$ , thus, $\sin\theta > \mu \cos \theta_{1}$ may be true
Since it is now clear that body B can also move upward, opt (d) is also rejected.

Question:5.14

Two billiard balls A and B, each of mass 50g and moving in opposite directions with speed of 5m s-1 each, collide and rebound with the same speed. If the collision lasts for $10^{-3} s$ , which of the following statements are true?
(a) The impulse imparted to each ball is 0.25 kg m s^-1 and the force on each ball is 250 N.
(b) The impulse imparted to each ball is 0.25 kg m s^-1 and the force exerted on each ball is 25 × 10^-5 N.
(c) The impulse imparted to each ball is 0.5 Ns.
(d) The impulse and the force on each ball are equal in magnitude and opposite in direction.

Answer:

The correct answer is:
(c) The impulse imparted to each ball is 0.5 Ns.
(d) The impulse and force on each ball are equal in magnitude and opposite in direction.
Let m be the mass of each ball, m = 0.05 kg
Let v be the speed of each ball, v = 5m/s
We know that the initial momentum of each ball will be
$\overrightarrow{p_{i}}=m\overrightarrow{v}$
$\overrightarrow{p_{i}}=(0.05)(5)=0.25kgms^{-1}$
= 0.25 N-s ……….. (i)
After the collision, on rebounding, the direction of the velocity of each ball is reversed; hence, the final momentum of each ball will be
$\overrightarrow{p}=m(-\overrightarrow{v})$
=- 0.25 N-s.
Hence, opt (d) is verified.
Thus, it is clear that impulse imparted to each ball is equal to change in momentum of each ball
$= p_{f} - p_{i}$
= - 0.25 – (0.25)
= -0.50 kg ms^-1
= -0.50 N-s
Here opt (c) is also verified.

Question:5.15

A body of mass 10 kg is acted upon by two perpendicular forces, 6N and 8N. The resultant acceleration of the body is
(a) 1 m $s^{-2}$ at an angle of $\tan^{-1}\left ( \frac{4}{3} \right )$ w.r.t. 6N force.
(b) 0.2 m $s^{-2}$ at an angle of $\tan^{-1}\left ( \frac{4}{3} \right )$ w.r.t. 6N force.
(c) 1 m $s^{-2}$ at an angle of $\tan^{-1}\left ( \frac{3}{4} \right )$ w.r.t. 8N force.
(d) 0.2 m $s^{-2}$ at an angle of $\tan^{-1}\left ( \frac{3}{4} \right )$ w.r.t. 8N force.

Answer:

The correct answer is:
(a) 1 m $s^{-2}$ at an angle of $\tan^{-1}\left ( \frac{4}{3} \right )$ w.r.t. 6N force.
(c) 1 m $s^{-2}$ at an angle of $\tan^{-1}\left ( \frac{3}{4} \right )$ w.r.t. 8N force.
Explanation:
Here we know that,
capture-36

m = 10kg, F₂ = 8N & F₁ = 6N
$R =\sqrt{ 8^{2}+6^{2}}$
$R =\sqrt{ 64+36}$
$R =\sqrt{ 100}$
Thus, R = 10 N
$F = ma \rightarrow a = \frac{F}{m} = \frac{R}{m} = \frac{10}{10} = 1ms^{-2}$ ………. (i)
$tan \theta_{1} =\frac{8}{6} = \frac{4}{3}=\theta_{1} = \tan^{-1}\left (\frac{4}{3} \right )$ ………… (ii)
$tan \theta_{2} =\frac{6}{8} = \frac{3}{4}=\theta_{2} = \tan^{-1}\left (\frac{3}{4} \right )$ ……….. (iii)
Hence, opt (a) is verified from (i) and (ii), whereas opt (c) is verified from (i) and (iii).
Since, acceleration is not equal to 0.2 ms^-2, opt (b) & (d) are rejected.

NCERT Exemplar Class 11 Physics Solutions Chapter 5 Very Short Answer

Question:5.16

A girl riding a bicycle along a straight road with a speed of 5 m s^-1 throws a stone of mass 0.5 kg which has a speed of 15 m s^-1 with respect to the ground along her direction of motion. The mass of the girl and bicycle is 50 kg. Does the speed of the bicycle change after the stone is thrown? What is the change in speed, if so?

Answer:

Here it is given that: m₁ = 50kg & m₂ = 0.5 kg
u₁ = 5 m/s forward & u₂ = 5 m/s forward
v₂ = 15 m/s forward
To find: v₁ = ?
By the law of conservation of momentum-
Initial momentum (i.e., girl, cycle & body) = Final momentum (girl, cycle & body)
$(50 + 0.5) \times 5 = 50 \times v_{1} + 0.5 \times 15$
$50.5 \times 5 - 7.5 = 50v_{1}$
$50v_{1} = 252.5 - 7.5 = 245$
Thus, $v_{1} = \frac{245}{50} = 4.9 m/s$
Thus, the speed of the cycle and the girl is reduced by 5 – 4.9 = 0.1 m/s.

Question:5.17

A person of mass 50 kg stands on a weighing scale on a lift. If the lift is descending with a downward acceleration of 9 m s^-2, what would be the reading of the weighing scale? (g = 10 ms² )

Answer:

The apparent weight would decrease on the weighing scale if the lift is descending with an acceleration ‘a’.
Thus, W’ = R = (mg – ma)
= m (g – a)
Due to reaction force, apparent weight by the lift on the weighing scale would be,
W’ = 50 (10 – 9) =50 N
Thus, the reading of the weighing scale would be
$\frac{R}{g}=\frac{50}{10}=5kg$

Question:5.18

The position time graph of a body of mass 2 kg is as given in Figure. What is the impulse on the body at t = 0 s and t = 4 s.

capture-37

Answer:

m = 2kg
Let v1 be the initial velocity, v₁ = 0, from the graph t ≥ 0 to t ≤ 4, viz., a straight line. Hence, the velocity of the body will be constant.
$v_{2} = \frac{3}{4} = 0.75 m/s$
At t ≥ 4,
The slope of the graph is zero; hence, the velocity v₃ = 0.
Now, impulse = $F .t = \frac{d\overrightarrow{p}}{dt} .dt = d\overrightarrow{p}$
We know that impulse is the change in momentum,
Impulse at t = 0;
= 2[0.75 – 0]
= 1.50 kg ms^-1
Impulse at t = 4;
= $m (v_{3} - v_{2})$
=2 [0 – 0.75]
=-1.50 kg ms^-1
Therefore, at t = 0, impulse increases by +1.5kg ms^-1 &
At t = 4, impulse decreases by 1.5 kg ms^-1

Question:5.19

A person driving a car suddenly applies the brakes on seeing a child on the road ahead. If he is not wearing seat belt, he falls forward and hits his head against the steering wheel. Why?

Answer:

Due to inertia of motion, if a person applies breaks suddenly, then that person’s lower part slows down rapidly, but the upper part continues to move in the same direction and with the same speed & thus he/she ends up hitting his/her head against the steering wheel of the car.

Question:5.2

The velocity of body mass 2kg as a function of t is given by $v(t)=2t\widehat{i}+ t^{2}\widehat{j}$ Find the momentum and the force acting on it, at t=2 s.

Answer:

Given data: m = 2kg

At t = 2s, $\overrightarrow{v }(2) = 2(2)\widehat{i} + (2)2\widehat{j}$
$\overrightarrow{v} (2) = 4\widehat{i}+ 4\widehat{j}$
Now, momentum $\overrightarrow{P }(2) = m\overrightarrow{v} (2)$
$\overrightarrow{P }(2) = 2[4\widehat{i} + 4\widehat{j}]$
$\overrightarrow{P }(2) = 8\widehat{i} + 8\widehat{j}kg ms^{-1}$
$\overrightarrow{F }(2) = m\overrightarrow{a}$
$\overrightarrow{F }(2) = m\overrightarrow{a}(2)$
$v(t)=2t\widehat{i}+ t^{2}\widehat{j}$
$\overrightarrow{a }(t) = \frac{d\overrightarrow{v}(t)}{dt} = 2\widehat{i} + 2t\widehat{j}$
$\overrightarrow{a }(2) = 2\widehat{i} + 2(2)\widehat{j}$
$= 2\widehat{i} + 4\widehat{j}$
Thus, $\overrightarrow{F} (2)= 2(2\widehat{i} + 4\widehat{j})$
$= 4\widehat{i} + 8\widehat{j}N$

Question:5.21

A block placed on a rough horizontal surface is pulled by a horizontal force F. Let f be the force applied by the rough surface on the block. Plot a graph of f versus F.

Answer:

Here,
A is the limiting frictional force & at B = kinetic frictional force.

Question:5.22

Why are porcelain objects wrapped in paper or straw before packing for transportation?

Answer:

During transportation, sudden jerks or falls takes place.
Porcelain objects are brittle and can break due to small jerks and falls, that’s why they are packed in paper or straw which provides them with more time to stop or change velocity during jerks and hence decreasing the acceleration. Hence the objects become safer since the force on them will be smaller.

Question:5.23

Why does a child feel more pain when she falls down on a hard cement floor, than when she falls on the soft muddy ground in the garden?

Answer:

From the formula ‘F = ma’, we can tell that if mass is kept constant then acceleration ‘a’ should be decreased for a system to decrease force.
Now, $a= \frac{v-u}{t}$ , for a falling body, initial velocity will be u and final velocity will be 0. The acceleration and force will decrease, if the time during hitting is increased. Pain acts on a person when he/she falls on a cemented hard floor as the time to stop after fall is very small, whereas, after falling on a soft floor he/she sinks in the ground and takes more time to stop.

Question:5.24

A woman throws an object of mass 500 g with a speed of 25 m s¹.
(a) What is the impulse imparted to the object?
(b) If the object hits a wall and rebounds with half the original speed, what is the change in momentum of the
object?

Answer:

(a) Given: m = 500g
= 0.5 kg,
u =0 & v = 25 m/s.
We know that,
$I = \Delta\overrightarrow{ P}$
$= m (\overrightarrow{v} - \overrightarrow{u} )$
= 0.5 (25 – 0)
= 12.5 N-s
(b) Here, m = 0.5 kg
u = +25 ms^-1 (forward)
v = $-\frac{25}{2}$ ms^-1 (as backward)
Thus, $\Delta p = m (v-u)$
$= 0.5 \left [-\frac{25}{2} - 25 \right ]$
= 0.5 [ -12.5 – 25]
$= 0.5 \times (-37.5)$
Therefore, $\Delta p = -18.75 kg ms^{-1}$
Therefore, $\Delta p$ or $\frac{\Delta p}{\Delta t}$ or the force is opposite to the initial velocity of the ball.

Question:5.25

Why are mountain roads generally made winding upwards rather than going straight up?

Answer:

When talking about inclined planes, for a body going upwards, the force of friction is $f_{s} = \mu N \cos \theta$ , where $\theta$ is the angle of inclination of a plane.
The force of friction is high if $\theta$ is small, and hence there is less chance of skidding.
The value of friction will be smaller for a road straight up the mountain since it will have a larger angle of inclination, and thus the chance of skidding will be more.

NCERT Exemplar Class 11 Physics Solutions Chapter 5 Short Answer

Question:5.26

A mass of 2 kg is suspended with thread AB (Figure). Thread CD of the same type is attached to the other end of 2 kg mass. Lower thread is pulled gradually, harder and harder in the downward direction so as to apply force on AB. Which of the threads will break and why?
capture-39

Answer:

The force on CD = force applied at D downward,
whereas the Force on AB = Force F along with force due to mass 2 kg downward.
Thus, the force on AB is more than fore at D by 2 kg, which will lead to the breakup of thread AB.
Hence, thread AB will break.

Question:5.27

In the above given problem if the lower thread is pulled with a jerk, what happens?

Answer:

If the thread CD is pulled with a jerk it will break since due to inertia of rest, pull on thread CD is not transmitted to thread AB by 2 kg mass.

Question:5.28

Two masses of 5 kg and 3 kg are suspended with help of massless inextensible strings as shown in the figure. Calculate $T_{1}$ and $T_{2}$ when whole system is going upwards with acceleration = 2 m/s².
capture-40

Answer:

We know that the whole system is going up with acceleration, $a = 2ms^{-2}$
In all parts of the string, tension is equal and opposite.
The forces acting on mass m₁ –
$T_{1} - T_{2} - m_{1}g = m_{1}a$
$T_{1} - T_{2} - 5g = 5a$
$T_{1} - T_{2} = 5g+ 5a$
$T_{1} - T_{2} = 5(9.8 + 2) = 5 \times 11.8$
$T_{1} - T_{2} = 59N$
The forces acting on mass m₂ –
$T_{2} - m_{2}g = m_{2}a$
$T_{2} = m_{2}(g+a)$
=3 (9.8 + 2)
=3(11.8)
= 35.4
Now, $T_{1} = T_{2} + 59$
= 35.4 + 59
Thus, $T_{1} = 94.4 N$

Question:5.29

Block A of weight 100 N rests on a frictionless inclined plane of slope angle $30^{\circ}$ (Figure). A flexible cord attached to A passes over a frictionless pulley and is connected to block B of weight W. Find the weight W for which the system is in equilibrium.
capture-42

Answer:

In this problem, the main concept used is of ‘on balance condition’, viz., if there is no motion, then there will be no frictional force or f=0.
During the equilibrium of A or B, mg sin $30^{\circ} = F$
$\frac{1}{2}mg = F$ …… (since mg is 100N)
Thus, $F= \frac{1}{2}(100)$
= 50
Hence, when B is at rest,
W = F = 50N.

Question:5.30

A bock of mass M is held against a rough vertical wall by pressing it with a finger. If the coefficient of friction between the block and the wall is $\mu$ and the acceleration due to gravity is g, calculate the minimum force required to be applied by the ginger to hold the block against the wall?

Answer:

Let M be the mass of the body which is held against the wall by force F. When the condition is balanced, F = N
capture-43

The minimum force required to hold the block against the wall at rest is $F = \frac{Mg}{ \mu}$

Question:5.31

A 100 kg gun fires a ball of 1 kg horizontally from a cliff of height 500 m. It falls on the ground at a distance of 400 m from the bottom of the cliff. Find the recoil velocity of the gun.

Answer:

Let u ms^-1 be the horizontal speed of the ball & its vertical component will be zero. If we consider the motion od the ball vertically downward then,
u = 0, s = h = 500m
g = 10sm^-2,
capture-44

We know that, $s = ut + \frac{1}{2} at^{2}$
$500 = a \times t + \frac{1}{2} \times 10t^{2}$
$t = \sqrt{100}$ = 10 s
Its horizontal range will be equal to $u \times 100$
$400 = v \times 10$
Thus, v = 40 m/s
$m_{b}$ is the mass of the ball
$M_{g}$ is the mass of the gun
$u_{b}$ is the initial velocity of the ball
$v_{g}$ is the final velocity of the gun
& $v_{b}$ is the final velocity of the ball
Thus $m_{g}(0) + M_{G}(0) = 1(40) + 100v_{G}$
Therefore, the recoil velocity of the gun will be $-\frac{40}{100}ms^{-1} = -\frac{2}{5} ms^{-1} = -0.4 ms^{-1}$ , viz., opposite to the speed of the ball.

Question:5.32

Figure shows (x, t), (y, t ) diagram of a particle moving in 2-dimensions.

capture-45
If the particle has a mass of 500g find the force (direction and magnitude) acting on the particle.

Answer:

By graph (a), $v_x = \frac{dx}{dt}= \frac{2}{2}$
= 1 ms^-1
$a_{x}=\frac{d^{2}x}{dt^{2}}=\frac{dv_{x}}{dt}=0$
By graph (b), $y = t^{2}$
Thus, $v_{y} = \frac{dy}{dt} = 2t$
$a_{y} = \frac{dv_{y}}{dt} = 2$
Thus, $F_{y} = ma_{y}$
m = 500g = 0.5 kg
$F_{y} = 0.5(2)$
= 1 N, towards Y-axis
$F_{x} = 0.5(0)$
= 0 N
$F = \sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{0^{2}+1^{2}}=\sqrt{1}$
Hence, F = 1 N, towards the Y-axis.

Question:5.33

A person in an elevator accelerating upwards with an acceleration of 2 m/s², tosses a coin vertically upwards with a speed of 20 m/s. After how much time will the coin fall back into his hand?

Answer:

Let ‘a’ be the upward acceleration of the elevator, a = 2m/s²
g = 10 ms^-2
Thus the net effective acceleration, viz., a’ is equal to (a+g)
Thus a’ = (2+10)
a’ = 12 ms^-1
Now, considering the effective motion of the coin, v = 0 & t = time taken by the coin to achieve maximum height
u = 20 ms^-1 and a’ = 12 ms^-2
thus, v=u + at here a = a’
0 = 20 – 12t
Therefore, $t = \frac{20}{12}s =\frac{5}{3} s$
Thus, time of ascend = time of descend
Therefore, after achieving maximum height, total time to return in hand will be equal to
$\frac{5}{3} + \frac{5}{3}$
= $\frac{10}{3}$
= $3\frac{1}{3}$ seconds.

NCERT Exemplar Class 11 Physics Solutions Chapter 5 Long Answer

Question:5.34

There are three forces $F_1, F_2$ , and $F_{3}$ acting on a body, all acting on a point P on the body. The body is found to move with uniform speed.
a) show that the forces are coplanar
b) show that the torque acting on the body about any point due to these three forces is zero

Answer:

(a) The acceleration of the body is zero since it is moving with uniform speed due to the action of the forces $F_1, F_2$ , and $F_{3}$ on a point on the body. It has no circular motion.
Since, F = ma, resultant force will be
$\overrightarrow{F_{1}}+\overrightarrow{F_{2}}+\overrightarrow{F_{3}}=0$
$\overrightarrow{F_{1}}+\overrightarrow{F_{2}}=-\overrightarrow{F_{3}} or \overrightarrow{F_{3}} = -(\overrightarrow{F_{1}}+\overrightarrow{F_{2}})$
Let us consider that the forces $\overrightarrow{F_1} and \overrightarrow{F_2}$ as well as the resultant of these forces, are in the same plane of the paper, but in $[-(\overrightarrow{F_{ 1}} + \overrightarrow{F_{ 2}})]$ it is in the same plane except for the direction which is rephrased.
$\overrightarrow{F_{3}}$ will be in the same plane since $\overrightarrow{F _{3}} = -(\overrightarrow{F_{ 1}} +\overrightarrow{ F_{ 2}})$ , hence $\overrightarrow{F_{ 1}} ,\overrightarrow{ F_{ 2}} and \overrightarrow{F _{3}}$ are coplanar.
(b) The resultant of $\overrightarrow{F_{ 1}} ,\overrightarrow{ F_{ 2}} and \overrightarrow{F _{3}}$ is zero, therefore, its torque= $r \times \overrightarrow{F} = 0$
$F=\overrightarrow{F_{ 1}} +\overrightarrow{ F_{ 2}} + \overrightarrow{F _{3}} =0$
Hence, the torque acting on the body at any point will be zero.

Question:5.35

When a body slides down from rest along a smooth inclined plane making an angle of $45^{\circ}$ with the horizontal, it takes time T. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance, it is seen to take time pT, where p is some number greater than 1. Calculate the coefficient of friction between the body and the rough plane.

Answer:

Since the body slides down from rest along a smooth plane viz. inclined at $45^{\circ}$ in time ‘T’.
u=0, s=s & t=T,
$a = g \sin 45^{\circ}=g/\sqrt{2}$

capture-46

Now, $s = ut + \frac{1}{2} at^{2}$
$s=\frac{gT^{2}}{2\sqrt{2}}$
The motion of the body along a rough plane,
u=0,
$s=\frac{gT^{2}}{2\sqrt{2}}$
………… (i)
$ma= mg \sin45^{\circ} - f$
$= mg\frac{ 1}{\sqrt{2}} - \mu N$
= $\frac{mg}{\sqrt{2}} - \mu mg \cos 45^{\circ}$

$ma =\frac{ mg}{\sqrt{2}} [1-\mu ]$
$a = \frac{g}{\sqrt{2}} (1- \mu )$
Now,
$s = \frac{g}{2\sqrt{2}} (1-\mu ) p^{2}T^{2}$ …….. (ii)

capture-47

It is given that in both cases the distance is equal
$\frac{gT^{2}}{2\sqrt{2}}=\frac{g}{2\sqrt{2}}\left ( 1- \mu \right )p^{2}T^{2}$

$1 = (1-\mu ) p^{2}$
$1 = p^{2} - \mu p^{2}$
$\mu p^{2} = p^{2} - 1$
$\mu =\left [ 1-\frac{1}{p^{2}} \right ]$

Question:5.36

Below figures show $(v_{x}, t)$ and $(v_{y}, t)$ diagrams for a body of unit mass. Find the force as a function of time.
capture-48

Answer:

From fig. (a),
V_x = 2t, for 0 < t < 1s
a_x = 2/1 = 2, for 0 < t < 1s
$V_{x} = 2(2-t)$ , for 1 < t < 2s
a_x= -2, for 1 < t < 2s
thus, F_x= ma (m = 1…..given)
= 1(2) =2, for 0 < t < 1s
& F_x = 1(-2)
= -2, for1 < t < 2s
From fig. (b),
$a_{y} = \frac{1}{1}$
=1 ms^-1, for 0 < t < 1
F_y = ma
= 1.1 = 1 unit, for 0 < t < 1
a_y = 0, for 1 < t
$F_{y}=1\times 0 =0$ units for 1 < t < 2s
$\overrightarrow{F} = \overrightarrow{F_{x}}\widehat{i} +\overrightarrow{ F_{y}}\widehat{j}$
$\overrightarrow{F} = 2\widehat{i} +4\widehat{j}$ for 0 < t < 1s
$\overrightarrow{F} =- 2\widehat{i} +0\widehat{j}$ , for 1 < t < 2s
Thus, $\overrightarrow{F}=-2\widehat{i}$ for 1 < t <2s
For more than 2 seconds
$a_{y}=0 , a_{x}=0$
$\therefore \overrightarrow{F}=0$

Question:5.37

A racing car travels on a track ABCDEFA. ABC is a circular arc of radius 2 R. CD and FA are straight paths of length R and DEF is a circular arc of radius R = 100 m. The coefficient of friction on the road is $\mu$ = 0.1. The maximum speed of the car is 50 m/s. Find the minimum time for completing one round.
capture-49

Answer:

To keep the car in circular motion centripetal force is provided by frictional force inward to centre O.
(i) Time is taken by the car to travel from A→B→C
Let s1 be the length of the path
Therefore, $s_{1} = \frac{3}{4}[2\pi (2R)]$
$= \frac{3}{4}(4\pi )(100)$
$=300 \pi m$
The maximum speed of the car along the circular path is $v_{1} = \sqrt{\mu rg}$
$v_{1} = \sqrt{0.1\times 2\times 100\times 10}=\sqrt{200}=10\sqrt{2}m/s$
= 14.14 m/s
Thus, $t_{1} =\frac{ s_{1}}{v_{1}} = \frac{300\pi}{14.14} = 66.62s$
(ii) Time taken by the car to travel from C to D & F to A
s₂ = CD + FA
= R + R
= 200 m
The car will travel with max. speed v₂ i.e., 50 m/s
Thus, $t_{2} =\frac{ s_{2}}{v_{2}}$
$\frac{200}{50}$
= 4 sec
(iii) Time taken for the path from D to E to F will be
$s_{3}=\frac{1}{4}2\pi R=\frac{1}{4}\times 2 \pi\times 100\times 50 \pi$
$v_{3}=\sqrt{\mu rg }=\sqrt{0.1\times R\times g}=\sqrt{0.1\times 100 \times 10}=10m/s$

$t_{3} = \frac{s_{3}}{v_{3}}$
$\frac{50 \pi}{10}$
$= 5\pi sec = 15.7 s$
Therefor the total time taken by car = $t_{1} +t_{2} + t_{3}$
= 66.62 + 4 +15.7
= 86.32s

Question:5.38

The displacement vector of a particle of mass m is given by
$r(t)=\widehat{i} A \cos \omega t + \widehat{j}B \sin \omega t$
a) show that the trajectory is an ellipse
b) show that $F = -m\omega ^{2}r$

Answer:

For plotting the graph (r-t) or trajectory we relate x & y coordinates.
(a)
$x = A \cos \omega t$ & $y = B \sin \omega t$
$\frac{x }{A}= \cos \omega t$ ……… (i)
$\frac{y}{B} = \sin \omega t$ ……… (ii)
On squaring and adding (i) & (ii), we get,
$\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=\cos ^{2}\omega t +\sin^{2} \omega t$
$\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1$

Here, we are getting the equation of an ellipse; hence, the trajectory is an ellipse.
(b) Given: $x = A \cos \omega t$
Therefore, $v_{x} = \frac{d_{x}}{dt} = -A\omega \sin \omega t$
& $a_{x} = \frac{dv_{x}}{dt} = -A\omega^{2} \cos \omega t$
Now, $y = B \sin \omega t$
Therefore $v_{y} = \frac{d_{y}}{dt }= B\omega \cos \omega t$
& $a_{y} = \frac{dv_{y}}{dt }= -B\omega^{2} \sin \omega t$
$a = a_{x}\widehat{i} + a_{y}\widehat{j}$
$= -\widehat{i} A\omega ^{2} \cos\omega t - \widehat{j} B\omega ^{2} \sin\omega t$
$= -\omega ^{2} [\widehat{i} A \cos \omega t + \widehat{j} B \sin \omega t]$
Therefore, $a = - \omega ^{2}\overrightarrow{r} (t)$
Thus, the force acting on particle= ma
$= -m \omega ^{2}\overrightarrow{r} (t)$

Question:5.39

A cricket bowler releases the ball in two different ways
a) giving it only horizontal velocity and
b) giving it horizontal velocity and a small downward velocity.
The speed v_s at the time of release is the same. Both are released at a height H from the ground. Which one will have greater speed when the ball hits the ground? Neglect air resistance.

Answer:

(a) $\frac{1}{2}v_{z}^{2} = gH$
Thus, $v_{z} =\sqrt{ 2gH}$
Speed of the ball at ground $=\sqrt{ v_{s}^{2}+v_{z}^{2} } = \sqrt{v_{s}^{2}+2gH}$
(b) Here as well, the total energy of the ball when it hits the ground is $\left [ \frac{1}{2} mv_{s}^{2} + mgH \right ]$
Hence the speed is same for (a) as well as (b).

Question:5.4

There are four forces acting at a point P produced by strings as shown in the figure, which is at rest. Find the forces F₁ and F₂.
capture-50

Answer:

a = 0 … (since the particles are at rest).
Hence, the resultant force & net components along X & Y axis will also be 0.
Let us resolve all the forces along X-axis,
F_x = 0,
capture-51

$F_{1} + 1 \cos 45^{\circ} - 2 \cos45^{\circ}= 0$
Or, $F_{1} - 1 \cos 45^{\circ} = 0$
$F_{1} = \cos 45^{\circ}$
$=\frac{ 1.414}{2}$
= 0.707 N
Now let us resolve all the forces along Y-axis
F_y = 0
$-F_{2} + 1 \cos 45^{\circ} - 2 \cos45^{\circ}= 0$
$-F_{2} =- 3 \cos 45^{\circ}$
= 3 (0.707)
= 2.121 N

Question:5.41

A rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is $\mu$ . Let the mass of the box be m.a) at what angle of inclination $\theta$ of the plane to the horizontal will the box just start to slide down the plane?
b) what is the force acting on the box down the plane, if the angle of inclination of the plane is increased to a > $\theta$
c) what is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed?
d) what is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration a.

Answer:

(a) By angle of repose $\mu = \tan \theta$ , as the box start to slide down the plane.
Therefore, $\theta = \tan^{-1} \mu$
(b) the angle of inclination of the plane with horizontal will slide down if angle $\alpha >\theta$ , as $\theta$ is the angle of repose.
Thus, the net downward force will be,
$F_{1} = mg \sin\alpha - f$
$= mg \sin\alpha -\mu N$
$= mg \sin\alpha -\mu mg \cos \alpha$
$= mg [ \sin \alpha - \mu \cos \alpha]$
(c) To move the box upward with uniform velocity or to keep it stationary,
$F_{2} - mg \sin \alpha - f = ma$
Or $F_{2} - mg \sin \alpha - \mu N =0$ ….. (since a=0)
$F_{2} = mg \sin \alpha - \mu N =0$
Therefore, $F_{2} = mg (\sin\alpha - \mu \cos \alpha)$
(d) Let F₃ be the force applied to move the box upward with acceleration a.
$F_{3} - mg \sin \alpha - \mu mg \cos \alpha = ma$
Thus, $F_{3} = mg( \sin \alpha - \mu \cos \alpha) + ma$

Question:5.42

A helicopter of mass 2000 kg rises with a vertical acceleration of 15 m/s². The total mass of the crew and passengers is 500 kg. Give the magnitude and direction of the
a) force on the floor of the helicopter by the crew and passengers
b) action of the rotor of the helicopter on the surrounding air
c) force on the helicopter due to the surrounding air

Answer:

Let, M be the mass of the helicopter, M = 2000 kg
m be the mass of the crew and passengers, m = 500 kg
& a be the acceleration of the helicopter with the passengers and crew, a = 15 ms^-2
(a) Force on floor of helicopter = apparent weight
= m (g + a)
= 500 (10 + 15)
Thus F₁= 500 (25)
= 12588N downward
(b)The action rotor of the helicopter which surrounds the air = Reaction force (by Newton’s 3^rd law),
Due to this, the helicopter rises up along with the crew and passengers with acceleration viz.,
15 ms^-1 = (M + m) (g + a).
Thus, the action of the rotor = (2000 + 500) (g + a).
Thus, F₂ = 2500 (10 + 15)
= 2500 (25)
= 62500 N downward.
(c) Let F₃ be the force acting on the helicopter by the reaction force of the surrounding air.
According to Newton’s 3^rd law
F₃ = -Action force
= -62500 N downward or F₃ = + 62500 N upward.

Students studying in Class 11 should have a detailed understanding of every chapter to succeed in his/her final exams and prepare for the board exams. NCERT Exemplar Class 11 Physics solutions chapter 5 PDF Download could be availed by the students through the website's link for a more precise explanation and resolution of queries.

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Main Subtopics in NCERT Exemplar Class 11 Physics Solutions Chapter 5 Laws of Motion

5.1 Introduction
5.2 Aristotle's fallacy
5.3 The law of inertia
5.4 Newton's first law of motion
5.5 Newton's second law of motion
5.6 Newton's third law of motion
5.7 Conservation of momentum
5.8 Equilibrium of a particle
5.9 Common forces in mechanics
5.10 Circular motion
5.11 Solving problems in mechanics.

What Will The Students Learn in NCERT Exemplar Class 11 Physics Solutions Chapter 5?

Students would think logically about Why the passengers fall backwards when a bus or train abruptly starts moving? Why do we lean to the right when a car makes a sharp left turn? Why does our weight in a lift fluctuate? NCERT Exemplar Class 11 Physics solutions chapter 5 provide the best understanding of the laws that govern motion. It describes the external force required to keep a body in motion and talks about the definition of inertia and how it affects our daily life. This chapter offers information on Newton's three laws of motion, with an introduction to Aristotle's fallacy. NCERT Exemplar Class 11 Physics chapter 5 covers the essential points on momentum, impulse, conservation of momentum, particles' equilibrium, and friction forces.

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NCERT Exemplar Class 11 Physics Solutions Chapter-Wise

Chapter 2 Units and Measurement

Chapter 3 Motion in a Straight Line

Chapter 4 Motion in a Plane

Chapter 6 Work, Energy, and Power

Chapter 7 Systems of Particles and Rotational Motion

Chapter 8 Gravitation

Chapter 9 Mechanical Properties of Solids

Chapter 10 Mechanical Properties of Fluids

Chapter 11 Thermal Properties of Matter

Chapter 12 Thermodynamics

Chapter 13 Kinetic Theory

Chapter 14 Oscillations

Chapter 15 Waves

Important Topics To Cover For Exams in NCERT Exemplar Solutions For Class 11 Physics Chapter 5 Laws of Motion

· NCERT Exemplar Class 11 Physics solutions chapter 5 introduces the students to the external forces that impact an object's motion by talking about Aristotle's fallacy and the concept of inertia. This chapter further describes in detail the three laws of motion, their definitions, and graphical derivations.

· This chapter covers the terms like inertia, momentum, force, and impulse. It talks about the inertia of rest, motion, and direction and highlights the concept of the conservation of momentum in launching a rocket. NCERT Exemplar Class 11 Physics chapter 5 allows a distinction in the different types of friction and describes a car's motion on an inclined plane.

· This chapter also talks about the variety of applications of these laws in many fields. NCERT Exemplar Class 11 Physics solutions chapter 5 provides an in-depth analysis of the motion of bodies in contact, the motion of bodies connected by strings, and a pulley Mass System using both a diagrammatic and graphical approach.

Check Class 11 Physics Chapter-wise Solutions

Chapter 1	Physical world
Chapter 2	Units and Measurement
Chapter 3	Motion in a straight line
Chapter 4	Motion in a Plane
Chapter 5	Laws of Motion
Chapter 6	Work, Energy and Power
Chapter 7	System of Particles and Rotational motion
Chapter 8	Gravitation
Chapter 9	Mechanical Properties of Solids
Chapter 10	Mechanical Properties of Fluids
Chapter 11	Thermal Properties of Matter
Chapter 12	Thermodynamics
Chapter 13	Kinetic Theory
Chapter 14	Oscillations
Chapter 15	Waves

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1. How can these solutions help?

These NCERT exemplar class 11 physics solutions chapter 5 can help you gain an insight into how these laws govern our everyday life and are an answer to many things happening around us. These solutions would also allow you to bag a good grade in 11th class exams and, in turn, clear your concepts for the board exams.

2. What are the essential topics of this chapter?

This essential topics of NCERT chapter laws of motion are the law of inertia, Newton's laws of motion, conservation of momentum, the equilibrium of a particle, and circular motion.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Exemplar Class 11 Physics Solutions Chapter 5 Laws of Motion

NCERT Exemplar Class 11 Physics Solutions Chapter 5 MCQI

NCERT Exemplar Class 11 Physics Solutions Chapter 5 MCQII

NCERT Exemplar Class 11 Physics Solutions Chapter 5 Very Short Answer

NCERT Exemplar Class 11 Physics Solutions Chapter 5 Short Answer

NCERT Exemplar Class 11 Physics Solutions Chapter 5 Long Answer

Main Subtopics in NCERT Exemplar Class 11 Physics Solutions Chapter 5 Laws of Motion

What Will The Students Learn in NCERT Exemplar Class 11 Physics Solutions Chapter 5?

NCERT Exemplar Class 11 Solutions

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Important Topics To Cover For Exams in NCERT Exemplar Solutions For Class 11 Physics Chapter 5 Laws of Motion

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