Question

A 144-g baseball moving 26 m/s strikes a stationary 5.25-kg brick resting on small rollers so it moves without significant friction. After hitting the brick, the baseball bounces straight back, and the brick moves forward at 1.05 m/s
(a) What is the baseball spped after the collision?
(b) Find the total kinetic energy before and after the collision.

Answer 1

Answer:

a. $v_{f1}=-12.28m/s$

b. $K_{Einitial} = 48.672J$,  $K_{E(final)} = 13.75 J$

Explanation:

Let the velocity of the ball after collision is v m/s, By the law of momentum conservation:

$p_i=p_f$

$m_1*u_1+m_2*u_2 = m_1*v_1 + m_2*v_2$

$0.144kg*26m/s + 5.25kg*0 = 0.144kg*v_{f1} + 5.25kg*1.05m/s$

Solve to vf1

$v_{f1}=\frac{0.144kg*26m/s-5.25kg*1.05m/s}{0.144kg}$

$v_{f1}=-12.28m/s$

(b)

$K_E(initial) = 1/2*m_1u_1^2$

$K_E = 1/2*0.144kg*(26m/s)^2$

$K_{Einitial} = 48.672J$

$K_{E(final)} = 1/2*m_1v_1^2 + 1/2*m_2v_2^2$

$K_{E(final)} = 1/2*0.144kg*(12.28m/s)^2 + 1/2*5.25 x (1.05m/s)^2$

$K_{E(final)} = 13.75 J$