A superball of mass 50 g is dropped from a height of 15 cm. (Since it is a super

Question

A superball of mass 50 g is dropped from a height of 15 cm. (Since it is a superball, assume that it rebounds from the surface with the same velocity with which it hit the surface.) If the superball was in contact with the table for 34 ms, calculate the average force exerted on the ball by the table. Show your work. (Hint: First calculate the momentum before and after hitting the table. Don?t forget the gravitational force.

N

The superball is now replaced with a clay ball that has the same mass and is dropped from the same height. What would be the average force exerted on the clay ball?

Explanation / Answer

let velocity of ball just before hitting the table is v

since, v2=u2+ 2as (equation of motion)

where,

v= final velocity ( v)

u = initail velocity = 0 (given)

a= accleration due to gravity = 9.81 m/s2

s= displacement = 15 cm = 0.15m

using equation

v= 0 + 2*9.81*0.15 = 2.943 m/s

since it is a super ball , hence after hitting the table the magnitude of velocity of the ball will remain same but its direction will be opposite

change in momentum = 2 m v

where m = mass of the ball = 50g = 0.05 kg

change in momentum = 2* 0.05 *2.943 = 0.2943 kg-m/s

force exerted (F) = change in momentum/ time

F = 0.2943/.034 = 8.65 N

in case of clay ball

change of momentum = mv = 0.14715 kg-m/s

F= 4.325 N

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