Question

A 22.3-g bullet moving at 1 000 m/s is fired through a one-kg block of wood emerging at a speed of 100 m/s. What is the change in the kinetic energy of the bullet-block system as a result of the collision assuming the block is free to move?

Answer 1

Answer:

-10837 J

Explanation:

The law of conservation of momentum states that the initial total momentum is equal to the final total momentum, so:

$p_i = p_f\\m u_b + M u_B = m v_b + M v_B$

where

m = 22.3 g = 0.0223 kg is the mass of the bullet

$u_b = 1000 m/s$ is the initial velocity of the bullet

M = 1 kg is the mass of the block

$u_B = 0$ is the initial velocity of the block

$v_b = 100 m/s$ is the final velocity of the bullet

$v_B$ is the final velocity of the block

Solving for $v_B$ we find

$v_B = \frac{m u_b-m v_b}{M}=\frac{(0.0223 kg)(1000 m/s)-(0.0223 kg)(100 m/s)}{1 kg}=20.1 m/s$

The total kinetic energy before the collision is:

$K_i = \frac{1}{2}mu_b^2 = \frac{1}{2}(0.0223 kg)(1000 m/s)^2=11,150 J$

And the total kinetic energy after the collision is:

$K_f = \frac{1}{2}mv_b^2 + \frac{1}{2}mv_B^2=\frac{1}{2}(0.0223 kg)(100 m/s)^2 + \frac{1}{2}(1 kg)(20.1 m/s)^2=313.5 J$

So, the change in kinetic energy is

$\Delta K = 313.5 - 11,150 J = -10,837 J$