Question

A point mass of $1kg$ collides elastically with a stationary with a point mass $5kg$. After their collision, the $1kg$ mass reverses its direction and moves with a speed of $2m/s$. Which of the following statement(s) is(are) correct for the system of these two masses?
A)Total momentum of the system is $3kgm/s$.
B)Momentum of $5kg$mass after collision is $4kgm/s$.
C)Kinetic energy of the centre of mass is $0.75J$.
D)Total kinetic energy of the system is $4J$.

Answer 1

Hint: Use the concept of conservation of linear momentum because there is no external force acting on the system of the two particles. Moreover, use the fact that the collision is purely elastic in nature.

Complete step by step solution:
Lets see every option one by one.

A)
As we can see, there is no external force on the system of two particles so we can use the law of conservation of momentum in this case. It means that the momentum of the system remains the same throughout the collision.
Let initial velocity of the $1kg$mass be $u$and final velocity of $5kg$mass be $v$
By the law of conservation of momentum,
$1(u) = 5(v) + 1( - 2)$
$u = 5v - 2$
Here, we take negative signs because it is given that the $1kg$mass reverses its direction.
By the formula of coefficient of restitution,
$e = \dfrac{{v - ( - 2)}}{u}$
But, the collision is elastic in nature which means that the value $e = 1$
$u = v + 2$
Putting the value of $u$in first equation
$5v - 2 = v + 2$
$v = 1m/s$
$u = 3m/s$
Initial momentum of the system is
$
  p = mu \\
  p = 1 \times 3 = 3kgm/s \\
 $
A)is correct

B)
Momentum of $5kg$mass after collision
$
  p = mv \\
  p = 5 \times 1 = 5kgm/s \\
 $
B)is correct

C)
Velocity of centre of mass,
 ${v_{cm}} = \dfrac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_1} + {m_2}}}$
${v_{cm}} = \dfrac{{1 \times 3 + 5 \times 0}}{{1 + 5}}$
${v_{cm}} = 0.5m/s$
Kinetic energy of centre of mass,
${K_{cm}} = \dfrac{1}{2} \times 6 \times {0.5^2}$
${K_{cm}} = 0.75J$
c)is correct.

D)
Since there is no force acting on the system, kinetic energy of the system is conserved. So we can find the kinetic energy of the system just by calculating the initial kinetic energy of the first particle.
${K_{system}} = \dfrac{1}{2}(1){(3)^2}$

${K_{system}} = 4.5J$
D) is incorrect.
So the correct options are A,B and C.

Note: Law of conservation of momentum is applicable here because the only forces in the system are in internal forces due to the collision of the two masses. These two forces are equal and opposite to each other therefore they cancel each other and net force on the system will be zero.