Question

A slab of clay with a mass of 0.47 kg and velocity of 8 m/s hits a stationary toy car of mass 0.56 kg. If the clay hits and sticks to the car, what is the velocity of the clay and car after impact?

Answer 1

Given:

The mass of the clay is

$m_1=0.47\text{ kg}$

The mass of the car is

$m_2=0.56\text{ kg}$

The initial velocity of the clay is

$v_1=8\text{ m/s}$

The initial velocity of the toy car is

$v_2=0\text{ m/s}$

Required: velocity after the impact

Explanation:

when a clay of mass m collides with a toy car then both of them move with the same velocity.

we will apply the momentum conservation here

that is given by

momentum before the impact= momentum after the impact

$m_1v_1+m_2v_2=(m_1+m_2)v$

Plugging all the values in the above relation, we get

$\begin{gathered} 0.47\text{ kg}\times8\text{ m/s+0.56 kg}\times0=(0.47\text{ kg+0.56 kg\rparen}\times v \\ v=\frac{3.76}{1.03} \\ v=3.65\text{ m/s} \end{gathered}$

Thus, the final velocity of clay and car is

$3.65\text{ m/s}$