Question

A point mass of 1kg collides elastically with a stationary point mass of 5kg. After their collision, the 1kg mass reverses its direction and moves with a speed of $2m{s^{ - 1}}$. Which of the following statement(s) is/are correct for the system of these two masses?
$A.Total{\text{ momentum of the system is 3kgm}}{{\text{s}}^{ - 1}}. \\\ B.Momentum{\text{ of 5kg mass after collision is 4kgm}}{{\text{s}}^{ - 1}}. \\\ C.Kinetic{\text{ energy of the center of mass is 0}}{\text{.75J}}{\text{.}} \\\ {\text{D}}{\text{.To kinetic energy of the system is 4J}}{\text{.}} \\\$

Answer 1

Hint- In order to solve this question, firstly we will understand the concept of conservation of linear momentum and apply it to find the total momentum of the system. Then we will find the velocity of center of mass i.e. ${V_{cm}} = \sum {\dfrac{{mV}}{M}}$ to get the required answer.

Formula used-
1. ${V_{cm}} = \sum {\dfrac{{mV}}{M}}$
2. $K.E = \dfrac{1}{2}m{V^2}$

Complete step-by-step answer:
Linear momentum conservation- In physics it is defined as a body or system of moving bodies that retains its total momentum, the product of mass and vector velocity. In other words the total momentum of a closed system is conserved,unless an external force is applied to it.
Applying the conservation of linear momentum,
The total momentum, before collision and after collision, equals the sum of the object’s individual momentum.
For each object, momentum is the product of its mass and its velocity which is measured in kilogram meters per second.
So using the above condition where before collision, m = 1kg of a point collides elastically with a stationary point mass of 5kg.
We get-
$1 \times v + 5 \times 0 = - 1 \times 2{v_2}.......\left( 1 \right)$
As collision is elastic,
${V_2} + 2 = V$
${V_2} = 1m{s^{ - 1}}$(Given)
$V = 3m{s^{ - 1}}$
Total momentum of the system$= 3kgm{s^{ - 1}}$.
Also velocity of center of mass is calculated by${V_{cm}} = \sum {\dfrac{{mV}}{M}}$
Where m is mass of point
V = speed
M is total mass
So we get-
${V_{cm}} = \dfrac{{1 \times V + 0}}{{1 + 5}}$
${V_{cm}} = \dfrac{1}{2}m/s$
Also $K.E = \dfrac{1}{2}m{V^2}$
So $K.E = \dfrac{1}{2}\left( {1 + 2} \right){V^2}cm = 0.75J$
Therefore, we conclude that kinetic energy of the center of mass is 0.75J and the total momentum of the system is $3kgm{s^{ - 1}}$.
Hence, option A and C are correct.

Note- While solving this question, we must know that after the collision, when the objects remain joined, they will move together with their combined momentum and if the objects move in opposite directions before the collision, the opposing velocities will partially cancel one another out.