Calculating Average Force on a Ball Hitting a Hard Surface

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To calculate the average force exerted by a hard surface on a ball weighing 10g that hits and rebounds at 5m/s, the correct approach involves using the formula F = M(v2 - v1)/t. The initial velocity is 5m/s downward, and the final velocity is 5m/s upward after rebounding, resulting in a total change in velocity of 10m/s. Substituting the values into the equation with time as 0.01 seconds yields an average force of 100N. The error in the initial calculation stemmed from incorrectly setting the final velocity to zero instead of recognizing the rebound speed. The correct average force is thus 100N.
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Homework Statement


A ball weighing 10g hit s ahard surface vertically with a speed of 5m/s and rebounds with the same speed.The ball remains in contact with the surface for 0.01 sec. The average force exerteed by the surface on the ball is :a) 100N b)10N c)1N d) 0.1N

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The Attempt at a Solution


using F= M(v2-v1)/t. I subs the values and converted g to kg.I also took the initial velocity as 5m/s and final as zero, time =0.01 sec . But I am getting 5 N. Where is my understanding flawed?
 
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takando12 said:
rebounds with the same speed

takando12 said:
final as zero
Compare those two statements.
 
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oh...initial and final are same. 5-(-5)=10. Stupid me.Thank you sir.
 
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