Question

A steel ball of mass 2m suffers one-dimensional elastic collision with a row of three steel balls, each of mass m. If mass 2m has collided with velocity v and the three balls numbered 1, 2, 3 were initially at rest, then what will happen after the collision?
(This question has multiple correct options)
${\text{A}}{\text{.}}$ Balls 1, 2 and 3 would start moving to right, each with velocity $\dfrac{{\text{v}}}{3}$
${\text{B}}{\text{.}}$ Balls 2 and 3 would start moving to right, each with velocity $\dfrac{{\text{v}}}{2}$
${\text{C}}{\text{.}}$ Balls 2 and 3 would start moving to right, each with velocity v
${\text{D}}{\text{.}}$ Ball 1 and ball of mass 2m would remain at rest

Answer 1

Here, we will proceed by evaluating the values of the linear momentum and kinetic energy of the complete system before collision. Then, we will apply the principles of conservation of linear momentum and kinetic energy for all the options.
Formula used:
p = mv and K.E = $\dfrac{1}{2}{\text{m}}{{\text{v}}^2}$.

Complete answer:
As shown in figure, the ball of mass 2m is moving with a velocity v towards a row of three balls each of mass m placed at rest.
It is given that the type of collision occurring is elastic.

As we know that the linear momentum p of the body having mass m and moving with velocity v is given by
p = mv $\to (1)$
The linear momentum of the whole system (including ball of mass 2m and three balls each of mass) before collision is given by
Total linear momentum of the system before collision = Linear momentum of ball of mass 2m moving with a velocity v + Linear momentum of the three balls (each of mass m and initially at rest) in a row
Using equation (1), the above equation can be written as
$\Rightarrow$Total linear momentum of the system before collision = (2m)v + (m+m+m)(0) = 2mv
Also we know that the kinetic energy K.E of the body of mass m and velocity v is given by
K.E = $\dfrac{1}{2}{\text{m}}{{\text{v}}^2}{\text{ }} \to {\text{(2)}}$
The kinetic energy of the whole system (including ball of mass 2m and three balls each of mass) before collision is given by

Total kinetic energy of the system before collision = kinetic energy of ball of mass 2m moving with a velocity v + kinetic energy of the three balls (each of mass m and initially at rest) in a row
Using equation (2), the above equation can be written as
$\Rightarrow$Total kinetic energy of the system before collision = $\dfrac{1}{2}\left( {{\text{2m}}} \right){{\text{v}}^2} + 3\left[ {\dfrac{1}{2}\left( {\text{m}} \right){{\left( 0 \right)}^2}} \right] = {\text{m}}{{\text{v}}^2}$
According to the principle of conservation of linear momentum,
Total linear momentum of the system after collision = Total linear momentum of the system before collision
$\Rightarrow$ Total linear momentum of the system after collision = 2mv $\to (3)$
According to the principle of conservation of kinetic energy,
Total kinetic energy of the system after collision = Total kinetic energy of the system before collision
$\Rightarrow$ Total kinetic energy of the system after collision = ${\text{m}}{{\text{v}}^2}{\text{ }} \to {\text{(4)}}$
So, the options which will satisfy both the equations (3) and (4) will be the correct answers.

${\text{A}}{\text{.}}$ Balls 1, 2 and 3 would start moving to right, each with velocity $\dfrac{{\text{v}}}{3}$
Here, velocity of ball of mass 2m = 0 (ball of mass 2m is at rest after collision)
Using equation (1), we can write
Total linear momentum of the system after collision = Linear momentum of ball of mass 2m at rest + Linear momentum of the three balls (each of mass m and velocity $\dfrac{{\text{v}}}{3}$) in a row
$\Rightarrow$ Total linear momentum of the system after collision = (2m)(0) + 3(m)($\dfrac{{\text{v}}}{3}$) = mv
Clearly, equation (3) is not satisfied. Therefore, option A is incorrect.

${\text{B}}{\text{.}}$ Balls 2 and 3 would start moving to right, each with velocity $\dfrac{{\text{v}}}{2}$
Here, velocity of ball of mass 2m = 0 (ball of mass 2m is at rest after collision) and velocity of ball numbered 1 of mass m = 0 (at rest)
Using equation (1), we can write
Total linear momentum of the system after collision = Linear momentum of ball of mass 2m at rest + Linear momentum of ball numbered 1 of mass m and placed at rest + Linear momentum of the two balls numbered 2 and 3 (each of mass m and velocity $\dfrac{{\text{v}}}{2}$) in a row
$\Rightarrow$ Total linear momentum of the system after collision = (2m)(0) + m(0) + 2(m)($\dfrac{{\text{v}}}{2}$) = mv
Clearly, equation (3) is not satisfied. Therefore, option B is also incorrect.

${\text{C}}{\text{.}}$ Balls 2 and 3 would start moving to right, each with velocity v
Here, velocity of ball of mass 2m = 0 (ball of mass 2m is at rest after collision) and velocity of ball numbered 1 of mass m = 0 (at rest)
Using equation (1), we can write
Total linear momentum of the system after collision = Linear momentum of ball of mass 2m at rest + Linear momentum of ball numbered 1 of mass m and placed at rest + Linear momentum of the two balls numbered 2 and 3 (each of mass m and velocity v) in a row
$\Rightarrow$ Total linear momentum of the system after collision = (2m)(0) + m(0) + 2(m)(v) = 2mv
Using equation (2), we can write
Total kinetic energy of the system after collision = kinetic energy of ball of mass 2m placed at rest + kinetic energy of the ball numbered 1 of mass m and placed at rest + Kinetic energies of the two balls numbered 2 and 3 (each of mass m and velocity v) in a row
$\Rightarrow$ Total kinetic energy of the system after collision = $\dfrac{1}{2}\left( {{\text{2m}}} \right){\left( 0 \right)^2} + \dfrac{1}{2}\left( {\text{m}} \right){\left( 0 \right)^2} + 2\left[ {\dfrac{1}{2}\left( {\text{m}} \right){{\left( {\text{v}} \right)}^2}} \right] = {\text{m}}{{\text{v}}^2}$
Clearly, both the equations (3) and (4) are satisfied. Therefore, option C is correct.

${\text{D}}{\text{.}}$ Ball 1 and ball of mass 2m would remain at rest (i.e., zero velocity)
Here, balls 2 and 3 can move in any direction with any velocity
Let us consider the case that both the balls 2 and 3 are moving in the right direction with velocity v each. This case is nothing but that given in option C which we are have already proved correct.

So, the correct answer is “Option C”.

Note:
In this particular problem, the collision which is happening is elastic collision and in any elastic collision, there occurs no loss in momentum as well as kinetic energy. That’s why we have used the principles of conservation of linear momentum and kinetic energy. Also, since momentum (or velocity) is a vector quantity, signs for direction need to be considered.