a ball weighing 10g hits a hard surface vertically With a speed of 5m/s & rebounds with the same speed. The ball remains in contact with the surface for 0.01s. What is the avg. force exerted by the ball on surface?
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Answered by
25
10 N as average force = ∆ P / ∆ t
where ∆P is the change in the momentum and ∆t is the time period..
We have
F=∆ P / ∆ t = (mV2 - mV1) / 0.01
F= 0.01 × { 5-(-5)} / 0.01
F = (0.01 × 10)/0.01
F= 10 N
where ∆P is the change in the momentum and ∆t is the time period..
We have
F=∆ P / ∆ t = (mV2 - mV1) / 0.01
F= 0.01 × { 5-(-5)} / 0.01
F = (0.01 × 10)/0.01
F= 10 N
lesy:
ye 5 and -5 kahan se aaya?
It's direction changes
Thnkx...
welcome
Answered by
6
Answer:
Explanation:
F=∆P/∆t
=Mv2-mv1/t
=m(v2-v1)/t
Now put the values,
0.01(5-(-5))/0.01. [ Rebounds the same speed =-5]
0.01×10/0.01= 10N
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