Question

A 3.0 kg black ball travels to the right at a speed of 1.7 m / s and collides with a stationary 6.0 kg red ball. After the collision, the black ball bounces and retraces its steps at a speed of 0.50 m / s. What is the speed in m / s of the red ball after the collision?

Answer 1

Let's write the law of conservation of linear momentum:

$$m_1v_{1i}+m_2v_{2i}=-m_1v_{1f}+m_2v_{2f}.$$

From this equation we can find the speed of the red ball, $v_{2f}$:

$$v_{2f}=\dfrac{m_1v_{1i}+m_2v_{2i}+m_1v_{1f}}{m_2},$$ $$v_{2f}=\dfrac{3.0\ kg\cdot 1.7\ \dfrac{m}{s}+6.0\ kg\cdot 0\ \dfrac{m}{s}+3.0\ kg\cdot 0.50\ \dfrac{m}{s}}{6.0\ kg}=1.1\ \dfrac{m}{s}.$$

The sign plus means that the red ball moves to the right.

Answer:

$v_{2f}=1.1\ \dfrac{m}{s}.$