Laws of Motion Test

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Laws of Motion Test - 5

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Question 1
4 / -1

Newton's second law of motion is

A
F = dp/dt

B
F = mv

C
F = mv²

D
F = m²v

Solution

Newton's second law of motion is
F = dp/dt = ma
Question 2
4 / -1

Which one of the following statemenls is not true about Newtons second law of motion

A
The second law of motion is consistent with the first law.

B
The second law of motion is a vector law.

C
The second law of motion is applicable to a single point particle.

D
The second law of motion is not a local law.

Solution

The second law of motion is a local law which means that force `F` at a point in space (location of the particle) at a certain instant of time is related to a at that instant. Acceleration here and now is determind by the force here and now, not by any history of the motion of the particle.
Question 3
4 / -1

The relation , cannot be deduced from Newton's second law, if

A
force depend on time

B
momentum depend on time

C
acceleration depend on time

D
mass depend on time

Solution

The relation , can only be deduced from Newton's second law, if mass remains constant with time. If mass depends on time then this relation cannot be deduced.
Question 4
4 / -1

A large force is acting on a body for a short time. The impulse imparted is equal to the change in:

A
acceleration

B
momemtum

C
energy

D
velocity

Solution

If a large force F acts for a short time dt the impulse imparted I is
I = Fdt = dp/dt dt
I = dp = change in momentum
Question 5
4 / -1

Which one of the following is not force?

A
Impulse

B
Tension

C
Thrust

D
Weight

Solution

Tension, thrust, weight are all common forces in mechanics whereas impulse is not a force.
Impulse = Force x Time Duration
Question 6
4 / -1

The motion of particle of mass m is given by y = ut + 1/2gt². The force acting on the particle is

A
mg

B
mu/t

C
2mg

D
2mu/t

Solution

Here y = ut + 1/2gt²
∴ v = dt/dy = u + gt
Acceleration, a = dt/dv = g
So, the net force acting on the particle is, F = ma = mg
Question 7
4 / -1

A constant force acting on a body of mass of 5 kg change its speed from 5ms^-1 to 10ms^-1 in 10s without changing the direction of motion, 'The force acting on the body is

A
1.5 N

B
2 N

C
2.5 N

D
5 N

Solution

Here,m = 5kg, M = 5ms^-1, v- 10ms^-1, t = 10s, Using v = u + at

As F = ma
∴ F = (5kg)(0.5ms⁻²) = 2.5N
Question 8
4 / -1

A bullet of mass 40g moving with a speed of 90 ms^-1 enters a heavy wooden block and is stopped after a distance of 60 cm. The average resistive force exerted by the block on the bullet is

A
180 N

B
220 N

C
270 N

D
320 N

Solution

Here, u = 90 ms⁻¹, v = 0
m = 40gm = 4/1000kg = 0.04 kg, s = 60 cm = 0.6 m
Using v²− u² = 2as
∴ (0)² − (90)² = 2a × 0.6

-ve sign shows the retradation.
∴ The average resistive force exerted by block on the bullet is
F = m × a = (0.04kg)(6750ms⁻²) = 270N
Question 9
4 / -1

A body of mass 5kg starts from the origin with an initial velocity . If a constant force N acts on the body, the time in which the y -component of the velocity becomes zero is

A
5s

B
20s

C
40s

D
80s

Solution

Here m = 5kg
∴ u_y = 40 m s^-1 and F_y = -5 N

= 1 m s^-1 v_y = 0
As vy = uy + ayt or 0 = 40-1 x t
∴ t = 40s
Question 10
4 / -1

A body under the action of a force acquiring an acceleration of 5ms⁻². The mass of the body is

A
2 kg

B
5 kg

C
4 kg

D
6 kg

Solution

Given: and a = 5ms⁻²
∴
Mass of the body is, m = F/a
= 10N / 5ms⁻²
= 2 kg
Question 11
4 / -1

A constant retarding force of -50 N is applied to a body of mass 10 kg moving initially with a speed of 10 m s^-1.The body comes to rest after

A
2 s

B
4 s

C
6 s

D
8 s

Solution

Here, F = -50 N (-ve sign for retardation) m=10 kg, u = 10 ms⁻¹, v = 0
As F = ma
∴ a =
Using v = u + at
∴ t =
= 2 s
Question 12
4 / -1

A body of mass 0.4 kg starting at origin at t = 0 with a speed of 10 m s^-1 in the positive x-axis direction is subjected to a constant F = 8 N towards negative x-axis. The position of the body after 25 s is

A
-6000 m

B
-8000 m

C
+4000 m

D
+7000 m

Solution

Mass of the particle, m = 0.4 kg
F = -8 N (minus sign for direction of force)
∴ Acceleration, a = F/m =
= -20 m s^-2
The position of the body at any time t is given by

The position of the body at t = 0 is 0, therefore x₀ = 0

Position of the body at t = 25 s
Here, u = 10ms⁻¹, a = −20ms⁻², t = 25s
∴ x = 10 × 25 + 1/2(−20)(25)²
= 250 − 6250 = −6000m
Question 13
4 / -1

If the force on a rocket, moving with a velocity of 300 m s^-1 is 210 N, then the rate of combustion of the fuel is

A
0.07 kgs^-1

B
1.4 kgs^-1

C
0.7 kgs^-1

D
10.7 kgs^-1

Solution

Force = d/dt(momemtum)
= d/dt(mv) = v(dm/dt) ⇒ 210 = 300(dm/dt)
rate of combustion, dm/dt = 210/300 = 0.7 kgs^-1
Question 14
4 / -1

A ball of mass m strikes a rigid wall with speed u and rebounds with the same speed. The impulse imparted to the ball by the wall is

A
2mu

B
mu

C
zero

D
-2mu

Solution

The situation is as shown in the figure.

p_initial= mu,p_final= −mu
Impulse imparted to the ball = change in momentum = p_initial= mu,p_final
= −mu − mu = −2mu
Question 15
4 / -1

A batsman hits back a ball of mass 0,15 kg straight in the direction of the bowler without changing its initial speed of 12 m s^-1. If the ball moves linearly, then the impulse imparted to the ball is

A
1.8 N s

B
2.8 N s

C
3.6 N s

D
4.2 N s

Solution

Change in momentum = 0.15 x 12 - (-0.15 x 12)
= 3.6 N s