Momentum and collisions

Question

A 0.045-kg bullet traveling at a speed of 456 m/s is fired vertically into a sponge with a mass of 0.0060kg that is initially at rest. Assuming that the bullet is embedded in the sponge, how high would the combined bullet and sponge rise after the perfectly inelastic collision?

Answer 1

You know momentum is always conserved, so m₁v_1i + m₂v_2i = mv_f where i stands for initial and f for final, and mass of bullet is m₁ and mass of sponge is m₂. Let m = m₁ + m₂. You should be able to determine the final velocity of the system (the velocity of the mass of bullet embedded in the sponge).

This final velocity is can be used as an initial velocity because you know that the system will travel upwards until its velocity is 0, and then will fall. You also know that the only force it experiences during this upward travel is gravity, so its acceleration is -9.8 m/s² (negative because its decelerating, meaning force is opposite of motion). Thus think of a kinematics equation which contains final velocity, initial velocity, the distance you are interested in, and acceleration.

You should be able to use v_f² = v_i²+2ad to determine the answer.