Physics Ncert Exemplar 2019 Solutions for Class 11 Science Physics Chapter 5 - Laws Of Motion

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Physics Ncert Exemplar 2019 Solutions for Class 11 Science Physics Chapter 5 Laws Of Motion are provided here with simple step-by-step explanations. These solutions for Laws Of Motion are extremely popular among class 11 Science students for Physics Laws Of Motion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Physics Ncert Exemplar 2019 Book of class 11 Science Physics Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Physics Ncert Exemplar 2019 Solutions. All Physics Ncert Exemplar 2019 Solutions for class 11 Science Physics are prepared by experts and are 100% accurate.

Page No 29:

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:

All parts of the body have the same velocity (magnitude and direction) and the velocity is constant in case of uniform translatory motion.

Hence, the correct answer is option (c).

Page No 29:

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:

As the scale is moving with uniform velocity, so its acceleration is zero. Thus, the force acting on the scale is zero. Also, the torque acting about its center of mass is zero.

Hence the correct answer is option (b).

Page No 30:

Question 5.3:

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

Answer:

Change in Momentum = final momentum - Initial momentum

Change in momentum

$P_{f} - P_{i} = m (v - u) = 0.15 ((- 3 \hat{i} - 4 \hat{j}) - (3 \hat{i} + 4 \hat{j})) P_{f} - P_{i} = - [0.9 \hat{i} + 1.2 \hat{j}]$

Hence, the correct answer is option (c).

Page No 30:

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 ms^–1
(c) 1.5 kg ms^–1
(d) 14 kg ms^–1

Answer:

The momentum transferred is given by:

$P_{f} - P_{i} = - [0.9 \hat{i} + 1.2 \hat{j}]$

Now, magnitude of the momentum transferred during the hit = $\sqrt{0 . 9^{2} + 1 . 2^{2}} = 1.5 kg {ms}^{- 1}$

Hence, the correct answer is option (c).

Page No 30:

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:

According to the law of conservation of momentum:

If no external force acts on a system (called isolated) of constant mass, the total momentum of the system remains constant with time.

Newton's Second Law:

$F = \frac{d P}{d t} 0 = \frac{d P}{d t} P = constant$
â€‹
Also, using Newton’s third law of motion, one can understand conservation of momentum in a collision between particles.

Hence, the correct answer is option (d).

Page No 30:

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:

According to Newton’s second law of motion only external forces can change linear momentum of the system.
The internal forces cannot change linear momentum of system under consideration.
The muscle force is the internal force, this cannot change the linear momentum of the player.
Hence, it is the friction force which changes the momentum of the player.
The rate of change of linear momentum of a body is equal to the external force applied on the body.
So, the external force must be in the direction of change in momentum as shown in the figure:

Initial and the final speed (v) of the player is same, so the magnitude of initial and final momentum $(|\vec{p_{1}}| = |\vec{p_{2}}| = m v)$ will be same. So, the of rate of change in momentum $(\vec{F} = \frac{\vec{p_{2}} - \vec{p_{1}}}{∆ t})$ is in the south-west direction.

Hence the correct answer is option (c)

Page No 30:

Question 5.7:

A body of mass 2 kg travels according to the law x(t) = pt + qt² + rt³where p = 3 ms⁻¹, q = 4 ms⁻² and r = 5 ms⁻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:

First we need to calculate the acceleration of the boy of mass 2 kg.
$Acceleration, a = \frac{d^{2} x}{d t^{2}} = (2 q + 6 r t) {ms}^{- 2} Acceleration, a = (8 + 30 t) {ms}^{- 2} Acceleration at time t = 2 seconds : a = 68 {ms}^{- 2}$

Force, F = $mass \times acceleration = 2 \times 68 = 136 N$

Hence, the correct answer is option (a).

Page No 31:

Question 5.8:

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

Answer:

Velocity along the x-axis becomes equal to zero, If the object has the velocity along y-axis.

Let's find the velocity of the mass along x-axis first

Acceleration is given as:

a = \frac{F}{m} = \{- \frac{3 \hat{i}}{5} + \frac{4 \hat{j}}{5}\} {ms}^{- 2} Using kinematic equation to find v_{x} v_{x} = u_{x} + a_{x} t

Putting v_xequal to zero gives:

v_{x} = u_{x} + a_{x} t - \frac{u_{x}}{a_{x}} = t \frac{6}{\frac{3}{5}} = t t = 10 s

Hence, the correct answer is option (b).

Page No 31:

Question 5.9:

A car of mass m starts from rest and acquires a velocity along east $v = v \hat{i} (v > 0)$ in two seconds. Assuming the car moves with uniform acceleration, the force exerted on the car is
(a) $\frac{m ν}{2}$ eastward and is exerted by the car engine.

(b) $\frac{m ν}{2}$ eastward and is due to the friction on the tyres exerted by the road.

(c) more than $\frac{m ν}{2}$ eastward exerted due to the engine and overcomes the friction of the road.

(d) $\frac{m ν}{2}$ exerted by the engine.

Answer:

Let us consider the eastward direction as x-axis.

A car is able to move due to friction acting between its tyres and the road.

The force of friction of the road on the tyres acts in the forward direction and is equal but in the opposite direction to the force of friction of the tyres on the road.

Mass of the car = m

As car starts from rest, â€‹its initial velocity, u = 0

Velocity acquired along east = v $\hat{i}$

Time interval (in which car acquired that velocity) t = 2 s.

As acceleration is uniform, so by applying the kinematic equation (v = u + at), we get:

$v = u + a t a = \frac{v}{2} \hat{i} Force F = \frac{m v}{2} \hat{i}$

The force acting on the car is along east with magnitude $\frac{m v}{2}$ . This is the frictional force acting â€‹on the tyres.

Hence, the correct answer is option (b).

Page No 31:

Question 5.10:

The motion of a particle of mass m is given by x = 0 for t < 0 s, x(t) = A sin 4π t for 0 < t < (1/4) s (A > 0), 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 –16π²Am.
(b) The particle is acted upon by on impulse of magnitude 4π²Am 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:

Finding the value of the acceleration a of the particle of mass m is

$x (t) = A \sin (4 π t) a (t) = - 16 π^{2} A \sin (4 π t)$

The force acting on the particle is given by mass times acceleration.

Force =

- 16 π^{2} A m \cos (4 π t)

t = \frac{1}{8} s

, Force =

- 16 π^{2} A m \sin (\frac{π}{2}) = - 16 π^{2} A m

Impulse

I = F \times t = - 16 π^{2} Am \times \frac{1}{4} = - 4 π^{2} Am

It can be clearly seen from the value of Force that it depends on the time t, So the particle is not acted upon by a constant force.

Hence, the correct options are (a), (b) and (d).

Page No 31:

Question 5.11:

In Fig. 5.1, 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 m/2 and of B is m. Which of the following statements are true?

(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:

Given mass of bodies, A and B are:
$m_{A} = \frac{m}{2} m_{B} = m$
The co-efficient of friction between the bodies B and A , $μ_{a} = 0.2$ .
the co-efficient of friction between the floor and the body B, $μ_{b} = 0.1$ .

When the two bodies move together, then acceleration of the combined body:

$a = \frac{F - f_{b}}{m_{A} + m_{B}} = \frac{F - f_{b}}{\frac{m}{2} + m} = \frac{2 (F - f_{b})}{3 m}$

Force on A:

$m_{A} a = \frac{m}{2} \times \frac{2 (F - f_{b})}{3 m} So force on A, F_{AB} = \frac{F - f_{b}}{3}$

If F_AB is equal to or smaller than f_b then body A will move along with body B.

$f_{b} = F_{AB} μ_{a} N_{A} = \frac{F - f_{b}}{3} 0.2 m_{A} g = \frac{F - f_{b}}{3} N_{A} = Reaction force by B on A f_{b} = μ_{b} N_{B} = μ_{b} (m_{A} + m_{B}) g [N_{B} = Normal reaction on B = (m_{A} + m_{B}) g] f_{b} = 0.1 \times (m_{A} + m_{B}) g f_{b} = 0.1 \times \frac{3}{2} m g = 0.15 m g$

The force (F) will be maximum when the blocks are moved together, then

$\frac{F_{\max} - f_{b}}{3} = μ m_{A} g = 0.2 \times \frac{m}{2} \times g = 0.1 m g \Rightarrow F_{\max} = 0.3 m g + f_{b} = 0.3 m g + 0.15 m g = 0.45 m g$

Similarly, we can find the minimum force required, so that both the blocks were moved together.

Hence, the maximum force up to which bodies will move together is F_max= 0.45mg. It verifies option (e). Hence option (e) is correct.

Both bodies can move together if F is less than or equal to 0.45mg.

So verifies the options (a) and (b) and rejects the option (c) as 0.5mg>0.45mg.

For option (d):
Minimum force for which A and B move together

$F_{\min} \geq f_{a} + f_{b} \Rightarrow F_{\min} \geq 0.2 \times \frac{m}{2} g + 0.15 m g F_{\min} \geq 0.25 m g$
Given force in option (d) 0.1mg <0.25mg
So body A and B will not move
â€‹Bodies A and B will remain in rest verifies option (d).

Hence, the correct options are (a), (b), (d) and (e).

â€‹

Page No 32:

Question 5.12:

Mass m₁ moves on a slope making an angle θ with the horizontal and is attached to mass m₂ by a string passing over a frictionless pulley as shown in Fig. 5.2. The co-efficient of friction between m₁and the sloping surface is μ.

Which of the following statements are true?
(a) If m₂ > m₁ sin θ, the body will move up the plane.
(b) If m₂ > m₁(sin θ + μ cos θ), the body will move up the plane.
(c) If m₂ < m₁ (sin θ + μ cos θ), the body will move up the plane.
(d) If m₂ < m₁ (sin θ − μ cos θ), the body will move down the plane.

Answer:

Consider the case when the body move up with acceleration a:

Let the tension in the string be T.
Normal reaction on the mass m₁,
N = m₁gcosθ
Friction force on the mass m₁,
$f = μ N = μ m_{1} g \cos θ So, T - m_{1} g \sin θ - f = m_{1} a . . . (1) Also, m_{2} g - T = m_{2} a . . . (2) Adding the above equations (1) and (2) we get : m_{2} g - (m_{1} g \sin θ + f) = (m_{1} + m_{2}) a \Rightarrow m_{2} g - (m_{1} g \sin θ + f) > 0 \Rightarrow m_{2} g - m_{1} g \sin θ - μ m_{1} g c o s θ > 0 (Friction, f = μ N = μ m_{1} g \cos θ) \Rightarrow m_{2} g > m_{1} g (\sin θ + μ \cos θ) \Rightarrow m_{2} > m_{1} (\sin θ + μ \cos θ)$
Hence, the option (b) is correct.

Similarly, if the body with mass m₁moves down and m₂ moves up then the friction is directed in the upward direction:

$- f + m_{1} g \sin θ > m_{2} g \Rightarrow - μ m_{1} g \cos θ + m_{1} g \sin θ > m_{2} g \Rightarrow m_{1} (- μ \cos θ + \sin θ) g > m_{2} g \Rightarrow m_{2} < m_{1} (\sin θ - μ \cos θ)$
So, option (d) is also correct.

Hence, the correct options are (b) and (d).

Page No 32:

Question 5.13:

In Fig. 5.3, a body A of mass m slides on plane inclined at angle θ₁ to the horizontal and μ₁ 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 θ₂ to the horizontal. Which of the following statements are true?

(a) A will never move up the plane.

(b) A will just start moving up the plane when $µ = \frac{\sin θ_{2} - \sin θ_{1}}{\cos θ_{1}}$ .

(c) For A to move up the plane, θ₂ must always be greater than θ₁.

(d) B will always slide down with constant speed.

Answer:

Let the tension in the string be T.

The normal reaction on block A and B is given by:
N₁ = mgcosθ₁
N₂ = mgcosθ₂
The friction is given by:
$f = μ N_{1} = μ m g \cos θ_{1}$
Now if the Body A moves up and B down the plane with acceleration a, then:

$T - (m g \sin θ_{1} + f) = m a . . . (1) m g \sin θ_{2} - T = m a . . . (2) Adding equation (1) and (2), m g \sin θ_{2} - (m g {sinθ}_{1} + f) = 2 m a At equilibrium : m g \sin θ_{1} + f = m g \sin θ_{2} m g \sin θ_{1} + μ m g \cos θ_{1} = m g \sin θ_{2} \Rightarrow μ = \frac{\sin θ_{2} - \sin θ_{1}}{{cosθ}_{1}}$

If we consider the case, when A moves upward and B moves downward.
$m g \sin θ_{2} - (m g {sinθ}_{1} + μ m g \cos θ_{1}) > 0$

$\frac{\sin θ_{2} - {sinθ}_{1}}{\cos θ_{1}} > μ (μ > 1) So, to meet the desired condition, θ_{2} > θ_{1}$

Hence, the correct options are (b) and (c).

Page No 32:

Question 5.14:

Two billiard balls A and B, each of mass 50g and moving in opposite directions with speed of 5 ms^–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 ms^–1 and the force on each ball is 250 N.
(b) The impulse imparted to each ball is 0.25 kg ms^–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:

We know that,
$Momentum, p = m v F o r c e, F = \frac{d p}{d t}$

Mass of balls A and B,
m_A = m_B = m = 0.005 kg each
The speed of the balls A and B are equal and is given by,
v_A = v_B = v = 5 m/s
Let us consider that the ball A is moving towards B, then the momentum of ball A is given by,
$p_{A} = m v$
$p_{A} = 0.005 \times 5 = 0.25 kgm / s$
Impulse imparted to each ball = Change in momentum of each ball
The momentum of ball A after the collision is given by,
$p'_{A} = - 0.25 kgm / s$ ' (Since, the ball rebound after collision with same speed)
So, the change in momentum of ball A is given by,
$p'_{A} - p_{A} = - 0.50 Ns$ ,
The magnitude of the impulse imparted is 0.50 Ns. So the statement (c) is correct i.e., the magnitude of impulse imparted by one ball due to collision with the other ball = 0.50 Ns.
The impulse and the force on each ball are equal in magnitude and opposite in direction.

Hence, the correct options are (c) and (d).

Page No 33:

Question 5.15:

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

Answer:

The resultant force is given by,
$R = \sqrt{8^{2} + 6^{2}} = 10 N$

So, the acceleration,
$a = \frac{F}{m} = \frac{10}{10} = 1 {ms}^{- 2}$

Now the angles can be calculated as:

${tanθ}_{1} = \frac{8}{6} \Rightarrow θ_{1} = \tan^{- 1} (\frac{4}{3}) And, {tanθ}_{2} = \frac{6}{8} \Rightarrow θ_{2} = \tan^{- 1} (\frac{3}{4})$

Hence, the correct options are (a) and (c).

Page No 33:

Question 5.16:

A girl riding a bicycle along a straight road with a speed of 5 ms^–1throws a stone of mass 0.5 kg which has a speed of 15 ms^–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:

Mass of girl and bicycle combined together is m₁ = 50 kg
Mass of the stone is m₂ = 0.50 kg
The initial velocity of the girl along with bicycle is given as u₁ = 5 m/s

Let us consider that the bicycle will move with a velocity v after the stone is thrown.

The initial momentum of the girl, the cycle and the stone combined together,

p_{i} = (50 + 0.5) \times 5 = 252.5 kg m / s

The final momentum is now given by,

p_{f} = 50 v + (0.5 \times 15) = 50 v + 7.5

Using the law of conservation of momentum here

p_{f} = p_{i} \Rightarrow 50 v + 7.5 = 252.5 \Rightarrow 50 v = 245 \Rightarrow v = \frac{245}{50} = 4.9 m / s

The change in speed of the cycle is (5 − 4.9) m/s = 0.1 m/s

Page No 33:

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 ms^–2, what would be the reading of the weighing scale?
(g = 10 ms^–2 )

Answer:

The weight of the person acts in downward direction,
mg = 500 N
The normal reaction R will act in the upward direction.
The lift is moving in the downwards direction with an acceleration of, a = 9 m s^-2
$m g - R = m a \Rightarrow 500 - R = 50 \times 9 \Rightarrow R = 50 N$
Hence, the weighing machine will show the reading 5 kg.

Page No 33:

Question 5.18:

The position time graph of a body of mass 2 kg is as given in Fig. 5.4. What is the impulse on the body at t = 0 s and t = 4 s.

Answer:

Impulse acting on the body at time t= 0 s

Impulse I = 0

Impulse acting on the body at t= 4s

Impulse I = change in momentum =

m \times (v - u)

I = - 2 \times \frac{3}{4} = - 1.5 {kgms}^{- 1}

Page No 34:

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, the upper part of the body remains to be in motion and it keeps on leaning in the forward direction even if the brakes are applied.

Also, the force of friction between the seat and the person is not enough to stop his motion in the forward direction.

So due to these reasons, the person hits his head on the steering wheel.

Page No 34:

Question 5.20:

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

Answer:

Momentum P is given by

$P (t) = m v (t) P (t) = 2 \times (2 t \hat{i} + t^{2} \hat{j}) P (t) = 4 t \hat{i} + 2 t^{2} \hat{j}$

At time t= 2s Momentum P is

$\vec{P} = 8 \hat{i} + 8 \hat{j}$

Force acting on the object is given by

$F (t) = \frac{d P (t)}{d t} = 4 \hat{i} + 4 t \hat{j}$

At t = 2 s Force is given by

$\vec{F} = 4 \hat{i} + 8 \hat{j}$

Page No 34:

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:

When a small force F is applied on a heavier box, it does not move. At this state, force of static friction f_s is equal to applied force F. On increasing applied force, the box does not move till the friction force reaches its limiting value, i.e., f_L = F. Further the box starts moving and in this situation a constant friction force called kinetic friction (f_k) starts acting on the box as shown in the graph:

Page No 34:

Question 5.22:

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

Answer:

To reduce the possibility of any damage to the porcelain objects they are wrapped in paper or straw before packing for transportation.
Porcelain objects are very fragile in nature. They get damaged easily even with small Force. So as to increase the time of impact during any halt while transportation they are wrapped in paper or straw. The basic idea is just to increase the time of impact on encountering Force.

The increase in time of impact reduces the effectiveness of the applied force.

Page No 34:

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:

More pain is felt by the child when the Force is large and vice- versa.

When the child falls on a soft muddy ground, the mud is redistributed on impact and time of contact of the child with the muddy ground increases. As we know from Newton's second law of motion that the Force is directly proportional to the rate of change of momentum, so the Force decreases with increase in time of contact.
So the force experienced by the child is more in case of cemented floor as compared to the muddy floor. That is why, 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.

Page No 34:

Question 5.24:

A woman throws an object of mass 500 g with a speed of 25 ms^-1.
(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) Impulse I = Change in Momentum
$I = 0.5 \times 25 = 1.25 k g m s^{- 1}$

(b) The object rebounds with half of the original velocity i.e. 12.5 ms^-1

$Change in momentum = m v - (- m \frac{v}{2}) Change in momentum = \frac{3 m v}{2} = 1.5 \times 0.5 \times 25 = 18.75 {kgms}^{- 1}$

Page No 34:

Question 5.25:

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

Answer:

Vehicles move on the road without skidding because of the presence of the force of friction between the road and the tyres of the vehicles .

This frictional force depends on the banking angle.

f = μ R = μ m g \cos θ

If θ is small, force of friction is high and there is less chance of skidding.

The road straight up would have a larger slope, So due to this reason the mountain roads are generally made winding upwards rather than going straight up.
â€‹

Page No 34:

Question 5.26:

A mass of 2 kg is suspended with thread AB (Fig. 5.5). 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?

Answer:

Thread AB will break , because force on the upper thread will be equal to sum of the weight of the body and the applied force.

Page No 34:

Question 5.27:

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

Answer:

Thread CD will break up if CD is pulled with the jerk because pull on thread CD is not transmitted from CD to AB instantaneously. Thus, before the mass can move, CD breaks.

Page No 34:

Question 5.28:

Two masses of 5 kg and 3 kg are suspended with help of massless inextensible strings as shown in Fig. 5.6. Calculate T₁ and T₂ when whole system is going upwards with acceleration = 2 ms² (use g = 9.8 ms^–2).

Answer:

Equation of motion can be written as

$T_{1} - (5 + 3) g = 8 \times 2 T_{1} = 16 + 78.4 = 94.4 N$

For T₂ the equation of motion can be written as

$T_{2} - 3 g = 3 a T_{2} = 3 (g + a) = 3 \times 11.8 = 35.4 N$

Page No 35:

Question 5.29:

Block A of weight 100 N rests on a frictionless inclined plane of slope angle 30° (Fig. 5.7). 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.

Answer:

From the diagram it is clear that the Force along the plane in the downward direction is equal to the force acting upwards along the plane

$mgsin 30^{o} = W 100 \times \frac{1}{2} = W W = 50 N$

Page No 35:

Question 5.30:

A block 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 μ and the acceleration due to gravity is g, calculate the minimum force required to be applied by the finger to hold the block against the wall ?

Answer:

If F is the force of the finger on the book and N is the normal reaction of the wall on the book, then:
F = N
The minimum upward frictional force needed to ensure that the book does not fall is Mg.
Frictional force = μN
Thus, the minimum value of the Force F is:
$F = N = \frac{M g}{μ}$ .

Page No 35:

Question 5.31:

100 kg gun fires a ball of 1kg 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. (acceleration due to gravity = 10 ms^–2).

Answer:

The vertical distance covered by the ball is given by the kinematic equation.

$y = u t + \frac{1}{2} g t^{2}$

From this equation time t can be calculated as $t = \sqrt{\frac{2 y}{g}}$

$t = \sqrt{\frac{2 \times 500}{10}} = 10 s$

Horizontal distance covered by the ball in this time interval is 400 m.

Horizontal velocity is given as $u_{x} = \frac{400}{10} = 40 {ms}^{- 1}$

The velocity with which ball is fired from the gun is 40 m s^-1.

Now using the law of conservation of momentum,
Final velocity of gun = $\frac{- 1 \times 40}{100} = - 0.4 {ms}^{- 1}$

So the recoil velocity comes out to be 0.4 m s^-1.

Page No 35:

Question 5.32:

Figure 5.8 shows (x, t), (y, t) diagram of a particle moving in 2-dimensions.

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

Answer:

From graph (a)

x = t

From graph (b)

y= t²

The acceleration a_x and a_y can be given as

a_x = 0

a_y = 2 ms^-2

Force would be along y direction only

F_{y} = 0.5 \times 2 = 1 N

Page No 35:

Question 5.33:

A person in an elevator accelerating upwards with an acceleration of 2 ms^–2, tosses a coin vertically upwards with a speed of 20 ms^–1. After how much time will the coin fall back into his hand? (g = 10 ms^–2).

Answer:

The sum of the acceleration due to gravity and the acceleration of the elevator = 10 + 2 = 12 ms^-2

Using the Kinematic equation,

$s = u t + \frac{1}{2} a t^{2} 0 = 20 t - (\frac{1}{2} \times 12 t^{2}) t = \frac{10}{3} s = 3.33 s$

Page No 35:

Question 5.34:

There are three forces F₁, F₂and F₃acting 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 body is moving with a uniform speed which means the net Force acting on the body is zero.

$\vec{F_{1}} + \vec{F_{2}} + \vec{F_{3}} = m a = m \times 0 = 0 \Rightarrow \vec{F_{1}} + \vec{F_{2}} = - \vec{F_{3}}$
Let $\vec{F_{1}} and \vec{F_{2}}$ be in the plane A containing both the forces.
Then F₁+ F₂ must be in the plane A.
Since, $\vec{F_{1}} + \vec{F_{2}} = - \vec{F_{3}}$ , so $\vec{F_{3}}$ is also in the plane A.
Hence, all the three forces $\vec{F_{1}}, \vec{F_{2}} and \vec{F_{3}}$ acting on a body are in the same plane, i.e., coplanar.

(b) Since all the forces pass through point P, so the torque of the forces about that point is zero.
Now consider torque about another point O.
Then torque about O,
= $\vec{OP} \times (\vec{F_{1}} + \vec{F_{2}} + \vec{F_{3}})$
= 0 $(Since, \vec{F_{1}} + \vec{F_{2}} + \vec{F_{3}} = 0)$
Hence, the torque acting on the body about any point due to these three forces is zero.
â€‹

Page No 36:

Question 5.35:

When a body slides down from rest along a smooth inclined plane making an angle of 45° 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 co-efficient of friction between the body and the rough plane.

Answer:

(a) Distance travelled by the body can be calculated using the kinematic equation.
When the surface is smooth i.e. there is no friction as shown in image:

Acceleration of the body along the plane in the downward direction is given by

a = g \sin θ

Distance travelled s is given as:

s = u t + \frac{1}{2} a t^{2} s = \frac{g \sin θ T^{2}}{2} T = \sqrt{\frac{2 s}{g \sin θ}}

T = \sqrt{\frac{2 s}{g \sin 45^{\circ}}} = \sqrt{\frac{2 \sqrt{2} s}{g}}

When the surface is rough i.e. there is friction as shown in image (b):

Time taken by the body is now given by pT.
acceleration in this case is given by:

m a = m g \sin θ - μ m g \cos θ \Rightarrow a' = \frac{(1 - μ) g}{\sqrt{2}} (Since, friction = μ N = μ m g \cos θ and θ = 45^{ο})

Hence, the time taken by the body to slide down the same displacement on the rough incline surface:

T' = \sqrt{\frac{2 s}{a'}} = \sqrt{\frac{2 \sqrt{2} s}{(1 - μ) g}} \Rightarrow p T = p \sqrt{\frac{2 \sqrt{2} s}{g}} = \sqrt{\frac{2 \sqrt{2} s}{(1 - μ) g}} \Rightarrow μ = 1 - \frac{1}{p^{2}}

Hence, the co-efficient of friction between the body and the rough plane is

μ = 1 - \frac{1}{p^{2}}

Page No 36:

Question 5.36:

Figure 5.9 shows (v_x, t), and (v_y, t) diagrams for a body of unit mass. Find the force as a function of time.

Answer:

From figure (a) we can conclude that

$v_{x} = 2 t f o r 0 < t \leq 1 = 2 (2 - t) f o r 1 < t < 2$

From figure (b) it can be concluded that

$v_{y} = t 0 < t < 1 s = 1 t > 1$

Acceleration can be calculated along x and y axis by taking the first derivative of velocity w.r.t time

Since the body have mass as 1 unit only, so the Force F can be written as

$F = \hat{i} + \hat{j} N 0 < t < 1 s = - 2 \hat{i} N 1 s < t < 2 s = 0 N t > 2 s$

Page No 36:

Question 5.37:

A racing car travels on a track (without banking) ABCDEFA (Fig. 5.10). 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 μ = 0.1. The maximum speed of the car is 50 ms^–1. Find the minimum time for completing one round.

Answer:

Here the whole circular track can be divided in two paths which are ABC and AEC respectively.

Along DEF, the velocity of the car along the track can be calculated as

v = \sqrt{g μ R} = \sqrt{100} = 10 m / s

The velocity of the car along ABC =

\sqrt{2 μ g R} = \sqrt{200} = 14.14 m / s

Time for DEF, t_{1} = \frac{π}{2} \times \frac{100}{10} = 5 π s Time for ABC, t_{2} = \frac{3 π}{2} \times \frac{200}{14.14} = \frac{300 π}{14.14} s

Time taken by the along the paths AF and DC is given by

t_{3} = \frac{2 R}{50} = \frac{200}{50} = 4 s

Total time taken by the car in one round trip is given by

T = t_{1} + t_{2} + t_{3} = 86.3 s

Page No 36:

Question 5.38:

The displacement vector of a particle of mass m is given by $r (t) = \hat{i} A \cos ω t + \hat{j} B \sin ω t$ .
(a) Show that the trajectory is an ellipse.
(b) Show that F = − mω²r.

Answer:

(a) $r (t) = \hat{i} A \cos ω t + \hat{j} B \sin ω t$ .

Here the value of the r vector is given and the x and y components of the trajectory of the particle are given by

x = A \sin (ω t) y = B \cos (ω t)

\frac{x^{2}}{A^{2}} + \frac{y^{2}}{B^{2}} = 1

The equation is the equation of the ellipse.

(b)

To calculate the acceleration of the particle:

\vec{a} = \frac{d^{2} \vec{r}}{d t^{2}} = - \hat{i} A ω^{2} \cos ω t - \hat{j} B ω^{2} \sin ω t \vec{a} = - ω^{2} (\hat{i} A \cos ω t + \hat{j} B \sin ω t) \vec{a} = - ω^{2} \vec{r}

\vec{F} = - {mω}^{2} \vec{r}

Page No 37:

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) When the ball is released with only horizontal velocity,
So, the initial velocity along the vertical direction is zero.
The initial velocity along horizontal direction is u_x= v_s
Since, there is no acceleration in horizontal direction, so the horizontal component of the final velocity will remain unchanged, i.e.,
$v_{x} = v_{s}$
Component of velocity along the vertical direction can be calculated using kinematic equation,
${(v_{y})}^{2} = {(u_{y})}^{2} + 2 g H = 0 + 2 g H \Rightarrow v_{y} = \sqrt{2 g H}$

Hence, the final velocity of the ball will be
$v_{f} = \sqrt{{(v_{x})}^{2} + {(v_{y})}^{2}} = \sqrt{{v_{s}}^{2} + 2 g H}$ ...(1)

(b) The ball is given horizontal velocity and a small downward velocity. The overall speed v_s at the time of release is the same in both the cases.
So initial kinetic energy of the ball is $\frac{1}{2} m {v_{s}}^{2}$
Let the final velocity of the ball when it hits the ground be v'_f
Then using the conservation of mechanical energy of the ball,
$K E_{i} + P E_{i} = K E_{f} + P E_{f} \Rightarrow \frac{1}{2} m {(v_{s})}^{2} + P E_{i} = \frac{1}{2} m {(v'_{f})}^{2} + P E_{f} \Rightarrow \frac{1}{2} m {(v_{s})}^{2} + (P E_{i} - P E_{f}) = \frac{1}{2} m {(v'_{f})}^{2} (Change in potential energy, P E_{i} - P E_{f} = m g H) \Rightarrow v'_{f} = \sqrt{{v_{s}}^{2} + 2 g H} . . . (2)$
From equation (1) and (2),
$v_{f} = v'_{f}$
Hence, the final velocity of the ball will be same in both the cases.

Page No 37:

Question 5.40:

There are four forces acting at a point P produced by strings as shown in Fig. 5.11, which is at rest. Find the forces F₁ and F₂.

Answer:

At equilibrium, net force on the block must be zero.
Resolving components along horizontal direction and adding them,
$F_{1} - 2 \cos (45^{\circ}) + \cos (45^{\circ}) = 0 \Rightarrow F_{1} = \cos (45^{\circ}) \Rightarrow F_{1} = \frac{1}{\sqrt{2}} N$
Resolving components along vertical direction,
$- F_{2} + \cos (45^{\circ}) + 2 \cos (45^{\circ}) = 0 \Rightarrow F_{2} = 3 \cos (45^{\circ}) \Rightarrow F_{2} = \frac{3}{\sqrt{2}} N$

Page No 37:

Question 5.41:

A rectangular box lies on a rough inclined surface. The co-efficient of friction between the surface and the box is μ. Let the mass of the box be m.
(a) At what angle of inclination θ 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 α > θ?
(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) From the definition of limiting friction,
the box will just start to slide down the plane when the angle of inclination is $θ = \tan^{- 1} (μ)$ .

(b)When angle of inclination α > θ:

Then the box will starts sliding down the incline plane.
the force acting on the box down the plane,
$F = mg \times sinα - f down the plane \Rightarrow F = mg \times sinα - μN \Rightarrow F = mg \times sinα - μmg \times cosα (Friction, f = μN = μmg \times cosα) \Rightarrow F = mg (sinα - μcosα) down the plane$

(c) When the box starts moving upwards, then friction force will starts acting in downward direction along the incline plane.
The force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed:
$F' = mg \times sinα + f \Rightarrow F' = mg \times sinα + μN \Rightarrow F' = mg \times sinα + μmg \times cosα \Rightarrow F' = mg (sinα + μcosα)$

(d) The force needed to be applied upwards along the plane to make the box move up the plane with acceleration a,
$F'' - (mgsinθ + μmgcosθ) = ma \Rightarrow F'' = ma + (mgsinθ + μmgcosθ)$

Page No 37:

Question 5.42:

A helicopter of mass 2000 kg rises with a vertical acceleration of 15 ms^–2. The total mass of the crew and passengers is 500 kg.
Give the magnitude and direction of the (g = 10 ms^–2)
(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:

Mass of the helicopter, M = 2000 kg
The total mass of the crew and passengers, m = 500 kg.
Acceleration, a = 15 ms^–2 upwards
and g = 10 ms^–2 downwards

(a) Force on the floor of the helicopter by the crew and passengers.
F = apparent weight of crew and passengers = $m (a + g)$
= 500 (10 + 15)
= 12500 N

(b) Action of rotor of helicopter on surrounding air is vertically downward direction, because helicopter rises on account of reaction of this force.
Thus action of rotor of helicopter on surrounding air,
F' =(m + M)× (a+g) = (2000 + 500) × (10 + 15)
â€‹ = 2500 × 25 = 62,500 N vertically downwards.

(c) Force on the helicopter due to surrounding air acts in upwards direction as a reaction force.
From Newton's third law, as action and reaction are equal and opposite, therefore Force of reaction,
F'' = 62,500 N vertically upwards.

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