Question

A smooth sphere of mass $M$ moving with velocity $u$ directly collides elastically with another sphere of mass m at rest. After collision, their final velocities are $V$ and $v$ respectively. The value of $v$ is:
A. $\dfrac{{2uM}}{m}$
B. $\dfrac{{2um}}{M}$
C. $\dfrac{{2u}}{{1 + \dfrac{m}{M}}}$
D. $\dfrac{{2u}}{{1 + \dfrac{M}{m}}}$

Answer 1

Hint: In this question, we are given the mass and the velocities of two spheres (after and before the collision). We have to find the value of $v$ (velocity of the second sphere after the collision). Firstly, apply Newton's collision law to calculate the value of $V$. Then, use the law of conservation of the momentum

Formula used:
Newton’s collision law,
$\dfrac{{{v_1} - {v_2}}}{{{u_1} - {u_2}}} = - e$
Where ${v_1},{v_2}$ are the velocities after the collision, ${u_1},{u_2}$ are velocities before the collision and $e$ is the coefficient of restitution
Momentum – Linear, translational, or simply momentum is the product of an object's mass and velocity in Newtonian mechanics.
$P = mv$
Here, $m$ and $v$ are equal to mass and speed of the body respectively.
Conservation of total momentum –
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\]
Here, ${m_1},{m_2}$ are the masses of the body, ${u_1},{u_2}$ is the initial velocities of the bodies and ${v_1},{v_2}$ are the final velocities.

Complete step by step solution:
Given that,
Mass of the sphere $1$, ${m_1} = M$
Mass of the sphere $2$, ${m_2} = m$
Initial velocity of the sphere $1$, ${u_1} = u$
Initial velocity of the sphere $2$, ${u_2} = 0$
Final velocity of the sphere $1$, ${v_1} = V$
Final velocity of the sphere $2$, ${v_2} = v$

Now, using the Newton’s collision law,
$\dfrac{{{v_1} - {v_2}}}{{{u_1} - {u_2}}} = - e$
$ \Rightarrow \dfrac{{V - v}}{{u - 0}} = - 1 \\ $
$ \Rightarrow V = v - u \\ $ …………………… (1)
The collision is elastic in nature. Therefore, the total momentum of before and after the collision will be conserved. Applying conservation of the momentum,
$\text{Total Momentum before collision = Total Momentum after collision}$
It will be, ${m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$
Put all the values in the above equation,
It implies that, $Mu = MV + mv$

Using equation (1), we get
$Mu = M\left( {v - u} \right) + mv$
$\Rightarrow Mu = Mv - Mu + mv$
$\Rightarrow 2Mu = v\left( {M + m} \right)$
$\Rightarrow v = \dfrac{{2Mu}}{{\left( {M + m} \right)}} \\ $
$\therefore v= \dfrac{{2u}}{{1 + \dfrac{m}{M}}}$

Hence, option C 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, kinetic energy and momentum 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. Because of the transfer of some kinetic energy to something else. The most likely culprits are thermal energy, acoustic energy, and material deformation.