Question

A ball of mass 100 g initially moving with velocity 20 m/s collides with a surface and stops within 0.1s. Average force exerted by the surface on the ball will be equal to?
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Answer 1

Explanation:

f=m*(v-u)/t

=.1*(0-20)/.1

= -20N

Answer 2

Average force exerted by the surface on the ball will be equal  20 N if A ball of mass 100 g initially moving with velocity 20 m/s collides with a surface and stops within 0.1s

Given:

• mass = 100 g = 0.1 kg       (1 kg = 1000 g)
• initial velocity = 20 m/s
• Stop in 0.1 sec

To Find:

• Average force exerted by the surface on the ball

Solution

Force x Time = Change in Momentum

Momentum = Mass * Velocity

Step 1:

Calculate Initial Momentum  m = 0.1 kg and u = 20 m/s

0.1 x 20  = 2 kgm/s

Step 2:

Calculate Final Momentum  m = 0.1 kg and v = 0 m/s ( Stops)

0,1 x 0  = 0 kgm/s

Step 3:

Calculate Change in Momentum

0 - 2  = - 2 kgm/s

Step 4:

Find Rate of Change in Momentum  as time = 0.1 sec

F = -2 /0.1  = -20N

As Equal and opposite force is Applied by the Surface hence

Average force exerted by the surface on the ball will be equal  20 N

Another Method:

• F = ma
• a = (v - u)/t
• a = (0 - 20)/(0.1)
• a = -200 m/s²
• F = 0.1 * (-200) = -20 N
• As Equal and opposite force is Applied by the Surface hence
• Average force exerted by the surface on the ball will be equal  20 N