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?
Answers 2
Explanation:
f=m*(v-u)/t
=.1*(0-20)/.1
= -20N
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Author:
commandom2lx
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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
- 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
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Author:
pixieyesm
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