A ball of mass 9m is dropped from rest from a height H = 5.0 meters above the gr

Question

A ball of mass 9m is dropped from rest from a height H = 5.0 meters above the ground, as shown above on the left. It undergoes a perfectly elastic collision with the ground and rebounds. At the instant that the ball rebounds, a small blob of clay of mass m is released from rest from the original height H, directly above the ball, as shown above on the right. The clay blob, which is descending, eventually collides with the ball, which is ascending. Assume that g = 10 meters/second^2, that air resistance is negligible, and that the collision process takes negligible time. Determine the speed of the ball immediately before it hits the ground. Determine the time after the release of the clay blob at which the collision takes place. Determine the height above the ground at which the collision takes place. Determine the speeds of the ball and the clay blob immediately before the collision. If the ball and the clay blob stick together on impact, what is the magnitude and direction of their velocity immediately after the collision?

Explanation / Answer

a) v2=u2+2gs (v=imediatly before it hits the ground )

   v=sqrt(2*10*5) = 10 m/s

b) collision happen at distace x from the ground

   s=ut+0.5gt2

for 9m ball

x=10t-5t2 -----1

for clay ball

(5-x)=5t2

from 1

5-10t+5t2=5t2

t=0.5 s

c)x=5-5*0.25 =3.75 m from the ground

d)v=u+at

v=10*0.5 = 5m/s

e)since the both objects stick together total momentum is conserved (up)

velocity of 9m before the collision v=10-10*0.5=5m/s

9m*5-m*5=10m*v

v=4 m/s (in up direction)

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