Question

A ball P of mass 2 kg undergoes an elastic collision with another ball Q at rest. After collision, ball P continues to move in its original direction with a speed one fourth of its original speed. What is the mass of ball Q?

Answer 1

Answer on Question #76288, Physics Mechanics Relativity

A ball $P$ of mass $2\,\mathrm{kg}$ undergoes an elastic collision with another ball $Q$ at rest. After collision, ball $P$ continues to move in its original direction with a speed one fourth of its original speed. What is the mass of ball $Q$?

Solution.

From the law of conservation of momentum

$$m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'$$ $$m_1 (v_1 - v_1') = m_2 (v_2' - v_2)$$

From the law of conservation of energy

$$\frac{m_1 v_1^2}{2} + \frac{m_2 v_2^2}{2} = \frac{m_1 v_1'^2}{2} + \frac{m_2 v_2'^2}{2}$$

or

$$m_1 (v_1^2 - v_1'^2) = m_2 (v_2'^2 - v_2^2)$$

or

$$m_1 (v_1 - v_1')(v_1 + v_1') = m_2 (v_2' - v_2)(v_2' + v_2)$$

Take advantage of the law of conservation of momentum

$$v_1 + v_1' = v_2' + v_2$$ $$v_1 + \frac{v_1}{4} = v_2' + 0$$

where

$$v_1' = \frac{v_1}{4},\ v_2 = 0$$ $$v_2' = \frac{5 v_1}{4}$$

Substitute the values obtained in the law of conservation of energy

$$m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'$$ $$m_1 v_1 + m_2 \cdot 0 = m_1 \frac{v_1}{4} + m_2 \frac{5 \cdot v_1}{4}$$ $$\frac{3 \cdot m_1 v_1}{4} = m_2 \frac{5 \cdot v_1}{4}$$ $$m_2 = \frac{3 m_1}{5} = \frac{3 \cdot 2\,\mathrm{kg}}{5} = 1.2\,\mathrm{kg}$$

Answer: The mass of ball $Q$ is $1.2\,\mathrm{kg}$

Answer provided by https://www.AssignmentExpert.com