Source: High school physics (Chinese)
Problem Sets:
Problem
On a smooth horizontal surface lies a wooden block of mass 400 g. A bullet of mass 50 g hits the block with a horizontal velocity of 200 m/s. The bullet passes through the block, and its exit velocity is 160 m/s.
- What is the final velocity of the block?
- How much mechanical energy is lost in this process?
[Q1] $V_f = 5$ m/s [Q2] $\Delta E = 355$ J
The system is the bullet and the block. Horizontal momentum is conserved. Let $m, v_i, v_f$ be the bullet's mass, initial velocity, and final velocity. Let $M, V_i, V_f$ be the block's mass, initial velocity, and final velocity. Here $V_i=0$.
[Q1] Conservation of momentum:
$$m v_i + M V_i = m v_f + M V_f$$ $$V_f = \frac{m(v_i - v_f)}{M}$$ $$V_f = \frac{0.05 \text{ kg} (200 \text{ m/s} - 160 \text{ m/s})}{0.4 \text{ kg}}$$[Q2] The loss in mechanical energy is the difference between initial and final kinetic energies: $\Delta E = K_i - K_f$.
$$K_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(0.05 \text{ kg})(200 \text{ m/s})^2$$ $$K_f = \frac{1}{2}mv_f^2 + \frac{1}{2}MV_f^2 = \frac{1}{2}(0.05 \text{ kg})(160 \text{ m/s})^2 + \frac{1}{2}(0.4 \text{ kg})(5 \text{ m/s})^2$$ $$\Delta E = K_i - K_f$$