Question

A 12.0-g bullet is fi red horizontally into a 100-g wooden block initially at rest on a horizontal surface. After impact, the block slides 7.5 m before coming to rest. If the coefficient of kinetic friction between block and surface is 0.650, what was the speed of the bullet immediately before impact?

Answer 1

Answer:

$v_1 = 91.3 m/s$

Explanation:

By energy conservation and work energy theorem we can say that after bullet hits the block, it will move on the rough floor and comes to rest

so here work done by frictional force = change in kinetic energy

so we know that

$W_f = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

$-\mu mg d = \frac{1}{2}m(v_f^2 - v_i^2)$

$-0.650(9.81)(7.5) = \frac{1}{2}(0 - v_i^2)$

$47.8 = \frac{1}{2}v^2$

$v = 9.78 m/s$

now by momentum conservation we have

$m_1 v_1 = (m_1 + m_2) v$

$12 v_1 = (100 + 12) 9.78$

$v_1 = 91.3 m/s$