Question

A bullet of mass m = 51 grams, traveling with a velocity v upwards, strikes the bottom of a ball of mass M = 5.6 kg which is resting in a hole in a table. After the collision, the ball, with the bullet embedded in it, rises up and returns to the table after 0.57 seconds. How fast was the bullet moving as it struck the ball?

Answer 1

Derivation of time of flight:

For the following situation   = 90^o

t_f = 2u/g [Using this equation in the problem,

Given t_f = 0.57

u = 0.57*9.81/2 = 2.8 m/s .

This is the velocity of both ball and bullet together = 2.8 m/s [your answer]

The bullets original velocity is = 310 m/s

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