The velocity-time graph of an object of mass 10 kg moving along a straight line. Class 9

The velocity-time graph of an object of mass 10 kg moving along a straight line. Class 9

Question 1.

The velocity-time graph of an object of mass 10 kg moving along a straight line is shown in Fig. Calculate the force acting on the object by using the graph. Class 9

Answer:

From fig., the velocity increases linearly from 0 to 30 m s^-1 in 8 seconds (straight line graph).

Acceleration = [latex]\frac{\text { Change in velocity }}{\text { Time }}[/latex]

= [latex]\frac{(30-0)}{8}[/latex] = 3.75 m s^-2

Using F = ma: F = 10 kg × 3.75 m s^-2 = 37.5 N

Question 2.

A bullet of mass 50 g moving with a speed of 100 m s^-1 enters a heavy stationary wooden block and stops after penetrating a distance of 50 cm. Estimate the stopping force acting on the bullet (assume that the bullet undergoes constant acceleration within the block). Class 9

Answer:

Given: m = 50 g = 0.05 kg

u = 100 ms^-1

v = 0 m s^-1,

s = 50 cm = 0.5 m

Using v² = u² + 2as:

0 = (100)² + 2 × a × 0.5 0

= 10000 + a a

= -10000 m s^-2

Force = ma

= 0.05 × 1000

= 500 N

(acting in the direction opposite to motion of the bullet)