Question

A 4-kg ball traveling westward at 25 m/s hits a 15-kg ball at rest. The 4-kg ball bounces east at 8.0 m/s. What is the speed and direction of the 15-kg ball? What is the impulse of the second ball?

Answer 1

Given:

The mass of the first ball is,

`m_1=4\text{ kg}`

The initial velocity of the first ball towards West is,

`u_1=25\text{ m/s}`

The mass of thr second ball is,

`m_2=15\text{ kg}`

the second object is initially at rest.

The final velocity of the first ball is,

`v_1=-8.0\text{ m/s}`

we are taking West as positive.

Applying momentum conservation principle we can write,

`m_1u_1+m_2*0=m_1v_1+m_2v_2`

Substituting the values we get,

`\begin{gathered} 4*25+0=4*(-8.0)+15* v_2 \\ v_2=(100+32)/(15) \\ v_2=8.8\text{ m/s} \end{gathered}`

THe final velocity of the second ball is towards East and the magnitude is 8.8 m/s.

The impulse of the Second ball is,

`\begin{gathered} I=m_2v_2-m_2*0 \\ =15*8.8 \\ =132\text{ kg.m/s} \end{gathered}`