Question

A ball of mass 'm' moving with speed 'u' undergoes a head-on elastic collision with a ball of mass 'nm' initially at rest. Find the fraction of the incident energy transferred to the second ball.
A. $\dfrac{n}{{1 + n}}$
B. $\dfrac{n}{{{{\left( {1 + n} \right)}^2}}}$
C. $\dfrac{{2n}}{{{{\left( {1 + n} \right)}^2}}}$
D. $\dfrac{{4n}}{{{{\left( {1 + n} \right)}^2}}}$

Answer 1

Hint: As we all know that when the collision between two bodies is elastic then there is no loss of kinetic energy or momentum takes place. Both the kinetic energies and momentum of two individual bodies are conserved.

Complete step by step answer:
As we know that according to the conservation of linear momentum, the momentum of the balls before the collision is equal to the momentum of the balls after the collision.
Let 1 and 2 be subscripts representing the balls 1 and 2. Therefore, we can say that,
Initial momentum of the balls =Final momentum of the balls
$m{u_1} + nm{u_2} = m{V_1} + nm{V_2}$
$ \Rightarrow mu = m{v_1} + nm{v_2}$ …… (I)
As we can see that it is given to us that the collision is elastic, hence the coefficient of restitution $e$ is 1.
We have studied that the coefficient of restitution $e$ is defined as,
$e = \dfrac{{{V_2} - {V_1}}}{{{u_1} - {u_2}}}$ …… (II)
Here, ${V_1},{V_2}$ are the final velocities of the first ball and second ball and ${u_1},{u_2}$ are the initial velocities of first ball and second ball.
Since the second ball is at rest initially, hence ${u_2} = 0$ and since the first ball is moving with velocity $u$ initially hence
${u_1} = u$.
Now we will be substituting ${u_1} = u$ and ${u_2} = 0$ , and $e = 1$ in equation (II), it becomes,
$\begin{array}{l}
1 = \dfrac{{{V_2} - {V_1}}}{u}\\
 \Rightarrow {V_1} = {V_2} - u
\end{array}$
Now we will substitute ${V_1} = {V_2} - u$ in equation (I) then it becomes,
$\begin{array}{l}
 \Rightarrow mu = m\left( {{V_2} - u} \right) + nm{V_2}\\
 \Rightarrow 2mu = m{V_2} + nm{V_2}\\
 \Rightarrow 2u = \left( {1 + n} \right){V_2}\\
 \Rightarrow {V_2} = \dfrac{{2u}}{{\left( {1 + n} \right)}}
\end{array}$
Now, we can write the fraction of kinetic energy transferred to the ball of mass nm as,
\[ \Rightarrow x = \dfrac{{K{E_2}}}{{K{E_i}}}\]
Here \[K{E_2}\] is the kinetic energy of the second ball after the collision and \[K{E_i}\]is the initial kinetic energy of the ball moving with velocity u and x is the fraction of energy transferred and as we all know that \[K{E_2}\] is the part of \[K{E_i}\] and hence the term \[\dfrac{{K{E_2}}}{{K{E_i}}}\] gives the fraction of energy.
We can now substitute \[K{E_2} = \dfrac{1}{2}nm{V_2}^2\] and \[K{E_i} = \dfrac{1}{2}m{u^2}\] to find the value of x.
$ \Rightarrow x = \dfrac{{\dfrac{1}{2}nm{V_2}^2}}{{\dfrac{1}{2}m{u^2}}}$ ……. (III)
Now we will substitute ${V_2} = \dfrac{{2u}}{{\left( {1 + n} \right)}}$ in equation (III) to find the fraction of energy transferred.
Hence it becomes,
\[ \Rightarrow x = \dfrac{{\dfrac{1}{2}nm{{\left( {\dfrac{{2u}}{{\left( {1 + n} \right)}}} \right)}^2}}}{{\dfrac{1}{2}m{u^2}}}\]
\[\therefore x = \dfrac{{4n}}{{{{\left( {1 + n} \right)}^2}}}\]

Therefore, fraction of incident energy transferred to the second ball is \[\dfrac{{4n}}{{{{\left( {1 + n} \right)}^2}}}\] and the correct option is (D).

Note:
Suppose in the above question, if the collision between the bodies were not elastic then the coefficient of restitution would have between 0 and 1. For a perfectly inelastic collision, the coefficient of restitution is zero.