Question

In space (no gravity or friction), you throw a ball with mass 0.1 kg at a target with mass 1 kg. you throw the ball at a speed of 3 m/s. when the ball impacts the target, it sticks to it and they drift off together. 1) how much energy is in the translational energy of the block+ball system after the collision?

Answer 1

It is not specified but I assume the target is initially at rest.

We can use the conservation of momentum to calculate the final velocity of the ball+block system. The initial momentum is the one of the ball, since the target is at rest:

`p_i = m_b v_b`
where
`m_b = 0.1 kg` is the mass of the ball and
`v_b = 3 m/s` is the initial velocity of the ball.

The final momentum is instead:

`p_f = (m_b + m_B)v_f`
where
`m_B=1 kg` is the mass of the block. For the conservation of momentum,

`p_i=p_f`
so we can find the final velocity of the ball+block system:

`m_B v_b = (m_b+m_B)v_f`

`v_f = (m_b v_b)/(m_b+m_B) = ((0.1 kg)(3 m/s))/(0.1 kg+1 kg) =0.27 m/s`

And the translational energy of the block+ball system after the collision is equal to its kinetic energy:

`K= (1)/(2)(m_b+m_B)v_f^2= (1)/(2)(0.1 kg+1 kg)(0.27 m/s)^2=0.04 J`