Question

A runaway 185 kg wheeled skip is rolling east along a road with a velocity of +2.2 m s−1. A 2560 kg car moving to the west with a velocity of −9.7 m s−1 hits the skip.

After the collision the skip is moving west with a velocity of −14.4 m s−1.

What is the velocity of the car after the collision? (in m s−1 to 2 s.f)

Answer 1

The law of conservation of momentum says

$$m_1\vec v_1+m_2\vec v_2=m_1\vec v_1'+m_2\vec v_2'$$

Hence

$$m_1v_1-m_2 v_2=-m_1 v_1'+m_2 v_2'$$

$$v_2'=\frac{m_1}{m_2}(v_1+v_1')-v_2$$

$$v_2'=\frac{185}{2560}(2.2+14.4)-9.7=-8.5\;\rm m/s$$

The velocity of the car after the collision is directed westward.