Question

Find the approximate speed of the second ball immediately after the impact of the heavy ball moving with speed $v$ and then collides with a tiny ball. It is given that the collision is elastic.
A.$v$
B.$2v$
C.${\displaystyle \frac{v}{2}}$
D.${\displaystyle \frac{v}{3}}$

Answer 1

Hint:By the use of conservation of momentum and then using the conservation of energy, and then solving those equations, we will get the expressions for the velocities of balls. The initial speed of the second ball is zero and the mass of the second ball is greater than the first ball.

Complete step by step answer:
First of all, we will consider all the values which are used for solving this problem. So –
Let the mass of heavy ball and tiny ball be the $m_1$ and $m_2$ respectively, initial velocities of heavy ball and tiny ball be $u_1$ and $u_2$ respectively and the final velocities be $v_1$ and $v_2$ respectively. Let the collision be in one – dimension.
According to the question, we know that –
Initial speed of second ball, $u_2=0$
Now, using the conservation of momentum, we get –
$m_1{v}_1+m_2{v}_2=m_1{u}_1$
By solving the value for $v_1$, we get –
$v_1={\displaystyle \frac{m_1{u}_1-m_2{v}_2}{m_1}}\cdots \left(1\right)$
Using the law of conservation of energy, we get –
$m_1{u}_1^2=m_1{v}_1^2+m_2{v}_2^2\dots \left(2\right)$
Using the value of $v_1$ from equation $\left(1\right)$ in equation $\left(2\right)$ -
$$\begin{gathered} m_1{u}_1^2=m_1{\left({\displaystyle \frac{m_1{u}_1-m_2{v}_2}{m_1}}\right)}^2+m_2{v}_2^2 \\ {m}_1{u}_1^2={\displaystyle \frac{1}{m_1}}\left(m_1^2{u}_1^2+m_2^2{v}_2^2-2m_1{u}_1{m}_2{v}_2\right)+m_2{v}_2^2 \\ \Rightarrow m_1{u}_1^2=m_1{u}_1^2+{\displaystyle \frac{m_2^2{v}_2^2}{m_1}}-2u_1{m}_2{v}_2+m_2{v}_2^2 \end{gathered}$$
Now, by transpositions, we get –
$$2u_1{m}_2{v}_2={\displaystyle \frac{m_2^2{v}_2^2}{m_1}}+m_2{v}_2^2$$
Cancelling $$m_2{v}_2$$ from both sides –
$2u_1={\displaystyle \frac{m_2{v}_2+m_1{v}_2}{m_1}}$
By solving for $v_2$ , we get –
$v_2={\displaystyle \frac{2m_1{u}_1}{m_1+m_2}}\cdots \left(3\right)$
Now, using the value of $v_2$ in the equation $\left(1\right)$ -
$v_1={\displaystyle \frac{m_1-m_2}{m_1+m_2}}{u}_1\cdots \left(4\right)$
According to the question, it is given that –
$$\begin{gathered} u_1=v \\ {u}_2=0 \end{gathered}$$
Let $v_2^{\prime }$ be the final velocity of second ball after collision –
Therefore, from equation $\left(3\right)$ and $\left(4\right)$, we get –
 $v_2^{\prime }={\displaystyle \frac{2m_1}{m_1+m_2}}{u}_1+{\displaystyle \frac{m_1-m_2}{m_1+m_2}}{u}_2\cdots \left(5\right)$
It is given that the second ball is heavy ball and the first ball is tiny –
$\therefore m_1>>m_2$
$$\begin{gathered} \because u_1=v \\ {u}_2=0 \end{gathered}$$
Putting these values in the equation $\left(5\right)$, we get –
$v_2^{\prime }=2v$
So, the correct answer is option B.

Note:The elastic collision can be defined as the collision in the system in which the total loss in kinetic energy of the system remains constant. In this collision, both momentum and kinetic energy are conserved.