Question

A
ball P of mass 0.25kg loses one-third of its velocity when it make a
head on a collision with an identical ball Q at rest .After collision,Q
moves off with a speed of 2m/s in the original direction of p.Calculate
the initial velocity of P?

Answer 1

QUESTION:

A ball $P$ of mass 0.25kg loses one-third of its velocity when it makes a head on a collision with an identical ball $Q$ at rest. After collision, $Q$ moves off with a speed of 2m/s in the original direction of $p$. Calculate the initial velocity of $P$?

SOLUTION:

According to the momentum conservation law, the momentum before collision is equal to the momentum after the collision:

$$m_P \upsilon_P = m_B \upsilon_B + m_P \upsilon_P'$$

$\upsilon_P' = \frac{2}{3} \upsilon_P$ — the velocity of a ball $P$ after the collision. Hence

$$m_P \upsilon_P = m_B \upsilon_B + m_P \frac{2}{3} \upsilon_P$$

According to the law of energy conservation, kinetic energy before the collision is equal to the kinetic energy after the collision. $K_{before} = \frac{m_P \upsilon_P^2}{2}$, $K_{after} = \frac{m_P \left(\frac{2}{3} \upsilon_P\right)^2}{2} + \frac{m_B \upsilon_B^2}{2}$. Hence

$$\frac{m_P \upsilon_P^2}{2} = \frac{m_P \left(\frac{2}{3} \upsilon_P\right)^2}{2} + \frac{m_B \upsilon_B^2}{2}$$ $$m_P \upsilon_P^2 = \frac{4}{9} m_P \upsilon_P^2 + m_B \upsilon_B^2$$

So,

$$\left\{ \begin{array}{l} m_P \upsilon_P = m_B \upsilon_B + m_P \frac{2}{3} \upsilon_P \\ m_P \upsilon_P^2 = \frac{4}{9} m_P \upsilon_P^2 + m_B \upsilon_B^2 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} m_P \left(\upsilon_P - \frac{2}{3} \upsilon_P\right) = m_B \upsilon_B \\ m_P \left(\upsilon_P^2 - \frac{4}{9} \upsilon_P^2\right) = m_B \upsilon_B^2 \end{array} \right. \Rightarrow$$ $$\begin{array}{l} \Rightarrow \left\{ \begin{array}{l} \frac{1}{3} m_P \upsilon_P = m_B \upsilon_B \\ \frac{5}{9} m_P \upsilon_P^2 = m_B \upsilon_B^2 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} \frac{1}{3} m_P \upsilon_P = m_B \upsilon_B \\ \frac{5}{9} m_P \upsilon_P^2 = m_B \upsilon_B^2 \end{array} \right. \Rightarrow \frac{\frac{5}{9} m_P \upsilon_P^2}{\frac{1}{3} m_P \upsilon_P} = \frac{m_B \upsilon_B^2}{m_B \upsilon_B} \Rightarrow \\ \Rightarrow \frac{5}{3} \upsilon_P = \upsilon_B \end{array}$$ $$\upsilon_P = \frac{3}{5} \upsilon_B$$ $$\upsilon_P = 1.2 \, \text{m/s}$$

ANSWER:

1.2 m/s