2 balls are dropped from a height of 6m. ball A bounces back up to height of 4m, whereas ball B bounces back up to 2m. which ball experience the larger impulse during it's collision with the floor?
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Given,
Mass of ball, m=0.5kg
Initial velocity is u=0.
Apply kinematic velocity, v2−u2=2as
Striking velocity, v=2gh=2×10×4=45ms−1
Ball reach to same height, hence rebound velocity magnitude is equal to striking velocity
Rebound velocity, v1=−45ms−1
Impulse, I=mΔV=m[v−(−v1)]=0.5(2×45)=45Ns−1
Force, F=ΔtmΔV=0.00245=4472.13N
Hence, impulse is 45Ns−1 and Force is 4472.13N
Given,
Mass of ball, m=0.5kg
Initial velocity is u=0.
Apply kinematic velocity, v2−u2=2as
Striking velocity, v=2gh=2×10×4=45ms−1
Ball reach to same height, hence rebound velocity magnitude is equal to striking velocity
Rebound velocity, v1=−45ms−1
Impulse, I=mΔV=m[v−(−v1)]=0.5(2×45)=45Ns−1
Force, F=ΔtmΔV=0.00245=4472.13N
Hence, impulse is 45Ns−1 and Force is 4472.13N
Answered by
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Answer:
Ball A experiences a larger impulse
This is so because more energy is required to bounce to 4m than 2m. This energy is provided by the impluse during collision. Therefore the ball which bounced to 4m (ball A) experiences larger impulse
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