# A ball with a mass of  3 kg is rolling at 16 m/s and elastically collides with a resting ball with a mass of  1 kg. What are the post-collision velocities of the balls?

May 21, 2016

#### Answer:

${\vec{v}}_{1}^{'} = 24 - 16 = 8 \text{ } \frac{m}{s}$

${\vec{v}}_{2}^{'} = 24 \text{ } \frac{m}{s}$

#### Explanation:

${\vec{P}}_{1} = {m}_{1} \cdot {\vec{v}}_{1} \text{ momentum of " m_1" before collision}$

${\vec{P}}_{2} = {m}_{2} \cdot {v}_{2} \text{ momentum of " m_2 " before collision}$

$\vec{\Sigma} {P}_{b} = {\vec{P}}_{1} + {\vec{P}}_{2} \text{ total momentum before collision}$

$\Sigma {\vec{P}}_{b} = {m}_{1} \cdot {\vec{v}}_{1} + {m}_{2} \cdot {\vec{v}}_{2}$

$\Sigma {\vec{P}}_{b} = 3.16 + 1 \cdot 0 = 48 \text{ } k g \cdot \frac{m}{s}$

$\text{post collision velocities :}$

${\vec{v}}_{1}^{'} = \frac{2 \cdot \Sigma {\vec{P}}_{b}}{{m}_{1} + {m}_{2}} - {\vec{v}}_{1}$

${\vec{v}}_{1}^{'} = \frac{2 \cdot 48}{3 + 1} - 16$

${\vec{v}}_{1}^{'} = \frac{96}{4} - 16$

${\vec{v}}_{1}^{'} = 24 - 16 = 8 \text{ } \frac{m}{s}$

${\vec{v}}_{2}^{'} = \frac{2 \cdot \Sigma {\vec{P}}_{b}}{{m}_{1} + {m}_{2}} - {\vec{v}}_{2}$

${\vec{v}}_{2}^{'} = \frac{2 \cdot 48}{3 + 1} - 0$

${\vec{v}}_{2}^{'} = \frac{96}{4}$

${\vec{v}}_{2}^{'} = 24 \text{ } \frac{m}{s}$