Question

A 0.22 kg piece of silly putty is thrown at a 1.2 kg wooden panel that is initially at the rest. The velocity of the silly putty before the collision was 18.5 m/s. The silly putty sticks to the wooden panel and they move as one after the collision. What is the velocity of the combined objects after the collision.

Answer 1

We can find the velocity of the combined objects after the collision from the Law of Conservation of Momentum:

$$m_pv_p+m_wv_w=(m_p+m_w)v,$$

here, $m_p=0.22\ kg$ is the mass of a piece of putty, $m_w=1.2\ kg$ is the mass of a wooden panel, $v_p=18.5\ \dfrac{m}{s}$ is the velocity of the piece of putty before the collision, $v_w=0\ \dfrac{m}{s}$ is the velocity of the wooden panel before the collision, $v$ is the velocity of the combined objects after the collision.

Then, from this equation we can find $v$:

$$v=\dfrac{m_pv_p}{m_p+m_w}=\dfrac{0.22\ kg\cdot 18.5\ \dfrac{m}{s}}{0.22\ kg+1.2\ kg}=2.86\ \dfrac{m}{s}.$$

Answer:

$v=2.86\ \dfrac{m}{s}.$