Question

A heavy steel ball of mass greater than \[1{\text{ }}kg\] moving with a speed of \[2{\text{ }}m/s\] collides head on with a stationary ping pong ball of mass less than \[0.1{\text{ }}g\]. The collision is elastic. After the collision the ping pong ball moves approximately with a speed ?
A. \[2{\text{ }}m/s\]
B. \[{\text{4 }}m/s\]
C. \[2 \times {10^4}{\text{ }}m/s\]
D. \[2 \times {10^3}{\text{ }}m/s\]

Answer 1

Hint:In this question, we are given that there are two balls i.e., steel ball and ping pong ball. Steel ball is heavier (\[1{\text{ }}kg\]) and the ping pong ball is the lighter ball (\[0.1{\text{ }}g\]). The ball with heavier mass collides at the speed of \[2{\text{ }}m/s\] with the lighter mass ball. To calculate the speed of a ping pong ball we have to apply the concept of conservation of linear momentum and kinetic energy.

Formula used:
Conservation of total momentum –
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\]
Conservation of total Kinetic energy –
\[\dfrac{1}{2}{m_1}{u_1}^2 + \dfrac{1}{2}{m_2}{u_2}^2 = \dfrac{1}{2}{m_1}{v_1}^2 + \dfrac{1}{2}{m_2}{v_2}^2\]
where $m$ and $v$ are the mass and speed of the body respectively.

Complete step by step solution:
Given that,
Mass of the steel ball, \[M = 1{\text{ }}kg\]
Initial Speed of the steel ball,\[v = 2{\text{ }}m/s\]
Initial speed of the ping pong ball, \[v' = 0\]
Mass of the ping pong ball, $m < 0.1g \approx m < 0.0001{\text{ }}kg$

Let, the final speed of the balls be ${v_1},{v_2}$ respectively.
Now, using the conservation law of momentum
\[Mv + mv' = M{v_{1\;}} + m{v_2}\]
\[\Rightarrow M\left( 2 \right) + m\left( 0 \right) = M{v_{1\;}} + m{v_2} \\ \]
\[\Rightarrow M(2-{v_1}) = m{v_2} - - - - - \left( 1 \right) \\ \]
Also, the nature of collision is elastic. Therefore, the kinetic energy will be conserved.
Using, conservation of kinetic energy
\[\dfrac{1}{2}M{v^2} + \dfrac{1}{2}m{\left( {v'} \right)^2} = \dfrac{1}{2}M{v_1}^2 + \dfrac{1}{2}m{v_2}^2 \\ \]
\[\Rightarrow \dfrac{1}{2}M{\left( 2 \right)^2} + \dfrac{1}{2}m{\left( 0 \right)^2} = \dfrac{1}{2}M{v_1}^2 + \dfrac{1}{2}m{v_2}^2 \\ \]
$ \Rightarrow M\left( {{2^2} - {v_1}^2} \right) = m{v_2}^2 - - - - - - \left( 2 \right)$

Dividing equation (1) by (2),
It will be \[\dfrac{{M(2-{v_1})}}{{M\left( {{2^2} - {v_1}^2} \right)}} = \dfrac{{m{v_2}}}{{m{v_2}^2}}\]
\[ \Rightarrow 2 + {v_1} = {v_2} \\ \]
Put \[2 + {v_1} = {v_2}\] in equation (1),
We get,
${v_1} = \dfrac{{2\left( {M - m} \right)}}{{\left( {M + m} \right)}} \\
\Rightarrow {v_2} = \dfrac{{4Mv}}{{\left( {M + m} \right)}}$
According to the question, $m$ is very small. Therefore, neglect $m$.
It implies that, ${v_1} = 2{\text{ }}m/s,{v_2} = 4{\text{ }}m/s$
After collision, the speed of the ping pong balls will be $4{\text{ }}m/s$.

Hence, option B is the correct answer.

Note: An elastic collision is one in which the system suffers no net loss of kinetic energy as a result of the collision. In elastic collisions, both momentum and kinetic energy are conserved. An inelastic collision occurs when there is a loss of kinetic energy. While the system's momentum is conserved in an inelastic collision, kinetic energy is not. This is due to the transfer of some kinetic energy to something else. The most likely culprits are thermal energy, acoustic energy, and material deformation.