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

a.
`v_(f1)=-12.28m/s`

b.
`K_(Einitial) = 48.672J`,
`K_(E(final)) = 13.75 J`

Step-by-step 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)=(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`