Falling ball collision

Question

A ball of mass 4m is dropped from rest from a height h = 10.0 m above the ground, as shown above. It undergoes a perfectly elastic collision with the ground and rebounds.

 

At the instant that the ball rebounds, a small piece 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, which is descending, eventually collides with the ball, which is ascending.

 

Assume that g = 10 m/s2 , air resistance is negligible, and the collision process takes negligible time.

 

a. Determine the speed of the ball immediately before it hits the ground.

 

b. Determine the time after the piece of clay is released at which the collision takes place.

 

c. Determine the height above the ground at which the collision takes place.

 

d. Determine the speeds of the ball and the piece of clay immediately before the collision.

 

e. If the ball and the clay stick together on impact, what is the magnitude and direction of their velocity immediately after the collision?

 

Answer 1

The ball falls with a speed that is increasing at the rate of 10m/s every second.

If it takes t seconds to reach the ground, its speed will be 10t m/s.

Its average speed is (0+10t)/2=5t. Since h=vt where v is its average speed t=h/v=10/5t, 5t²=10 and t=√2 seconds. The speed on reaching the ground is 10√2 m/s.

In a perfectly elastic collision with ground the ball will leave the ground at speed 10√2, but will be slowed down by gravity.

If the ball and clay collide after T seconds the ball will have reduced its upward speed by gT m/s so it will be 10√2-10T=10(√2-T) and its momentum will be 40m(√2-T) upwards. The clay will be moving at speed gT downwards with momentum mgT=10mT. Total momentum before collision is -40m(√2-T)+10mT.

The clay will have fallen a distance 5T² and the ball will have reached a height of 10T√2-5T².

Note that the clay and the ball fall at the same rate, so when we add the two distances together we get 10T√2=h=10 and T=1/√2=√2/2 seconds corresponding to a distance 10-5/2=7.5 m from the ground.

Immediately before the collision, the speed of the clay is gT=5√2=7.07m/s. The speed of the ball is 10√2-5√2=5√2=7.07m/s.

The combined downward momentum before the collision is mgT-40m(√2-T)=5m√2-20m√2=-15m√2. 

The clay-ball combo momentum after collision is (m+4m)V where V is the resultant velocity.

Because momentum is conserved, -15m√2=5mV so V=-3√2=4.24m/s. The resultant velocity is upward.