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Linear Momentum and Second Law of Motion

Question 1:
Figure shows an estimated force-time graph for a base ball struck by a bat from this curve determine impulse delivered to the ball.

1800 kg-m/sec

900 kg-m/sec

2700 kg-m/sec

None of these

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Solution:

Area bounded by force time graph with time axis represents impulse. Area of triangle in the figure is ⇒Impulse = ½×2 ×1800 = 1800 kg m/sec

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Question 2:
Acceleration of the block is

1 m/sec²

2 m/sec²

3 m/sec²

4 m/sec²

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Solution:

Net force acting on the object is 10 N.

so, acceleration a = F/m = 10 N/10 kg= 1 m/s²

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Question 3:
Acceleration of the block is :

1 m/s²

2 m/s²

3 m/s²

4 m/s²

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Solution:

Horizontal component of force is F cosθ = 20√2 cos 45° = 20√2 × 1/√2 = 20 N.

So, Horizontal acceleration = 20N / 10kg = 2m/s²

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Question 4:
A hammer of man 1 kg moving with a speed of 6 m/sec strikes a wall and comes to rest in 0.1 sec find average retarding force that stops the hammer :

40 N

50 N

60 N

None of these

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Solution:

According to Newtons second law of motion, Force is time rate of change in momentum.

So, F = change in momentum / time

F= m (vf-vi)/t= 1 kg (0-6)/0.1= -60 N

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Question 5:
A force F = (3ˆi + 4ˆj +12kˆ ) N produces an acceleration of 1 m/sec² in a body, find the mass of the body :

10 kg

13 kg

1 kg

12 kg

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Solution:

Given force is F = (3ˆi + 4ˆj +12kˆ ) N , magnitude of force is

⇒F²= (3²+4²+12²) so, F = 13 N

From Newton's second law of motion F= m.a

⇒ 13 N = m.1 so, m=13 kg

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Question 6:
A 5 kg block is resting on a frictionless plane, it is struck by a Jet, releasing water at the rate of 4 kg/sec emerging with a speed of 5 m/sec. Calculate the initial acceleration of block :

4 m/sec²

3 m/sec²

5 m/sec²

8 m/sec²

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Solution:

Force acting on the object is F = v.(dm/dt)= 5×4 = 20 N

so Acceleration is a= F/m = 20/5 = 4 m/s²

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Question 7:
A body of mass 1 kg is acted upon by two perpendicular forces of 8 N and 6 N, find the magnitude of acceleration :

10 m/sec²

6 m/sec²

8 m/sec²

20 m/sec²

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Solution:

Resultant force on the object will be F= (8² + 6²)½= 10 N

So, acceleration is a = F/m = 10/1 = 10 m/s²

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Question 8:
A small block slides without friction down an inclined plane starting from rest. Let Sn be the distance travelled from time t = n – 1 to t = n. Then Sn/Sn+1 is :

2n-1/2n

2n+1/2n-1

2n-1/2n+1

2n/2n+1

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Solution:

S_n = u + a/2(2n-1) = a/2(2n-1)

S_n+1 = u + a/2(2(n+1)-1) = a/2(2n+1)

S_n / S_n+1 = (2n-1)/(2n+1)

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Question 9:
Sand is poured on a conveyor belt at the rate of 2 kg/s. If belt is moving horizontally with velocity 4 m/s, then additional force required by engine to keep the belt moving with same constant velocity

8 N

10 N

12 N

14 N

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Solution:

The force required to keep the conveyor belt moving at a constant velocity is given by the rate of change of momentum.

Momentum of sand per second:

$\text{Force} = \frac{\text{dP}}{\text{dt}} = \frac{\text{dm}}{\text{dt}} \times v$

Given $\frac{dm}{dt} = 2$ kg/s and $v = 4$ m/s,

$F = 2 \times 4 = 8 \text{ N}$

Thus, the additional force required is 8 N.

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Question 10:
The position-time graph of a particle of mass 2 kg moving along x-axis is as shown in the figure. The magnitude of impulse on the particle at t = 2 s is

Zero

10 N-s

20 N-s

40 N-s

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Solution:

Impulse is given by the change in momentum:

$J = \Delta p = m \Delta v$

From the graph, we find velocity before and after $t = 2s$ :

Before $t = 2s$ :
$v_{\text{initial}} = \frac{x_2 - x_1}{t_2 - t_1} = \frac{0 - 20}{2 - 0} = -10 \text{ m/s}$
After $t = 2s$ :
$v_{\text{final}} = \frac{x_3 - x_2}{t_3 - t_2} = \frac{20 - 0}{4 - 2} = 10 \text{ m/s}$

Now, impulse:

$J = m (v_{final} - v_{initial})$

$J = m (v_{\text{final}} - v_{\text{initial}})$ $J = 2 \times (10 - (-10)) = 2 \times 20 = 40 \text{ Ns}$

Final Answer:

$\mathbf{40 \text{ Ns}}$

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