A point mass of 1kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1kg mass reverses its direction and moves with a speed of `2ms^-1`. Which of the following statements (s) is (are) correct for the system of these two masses?

Options

A. Total momentum of the sustem is 3 kg `m s^(-1)`
B. Momentum of 5 kg mass after collision is `4 kg m s^(-1)`
C. Kinetic energy of the centre of mass is 0.75 J
D. Total kinetic energy of the system is 4 J .

Answer

A, C

Explanation

In figure state of motion before collision is shown above the objects and state of motion after collision is shown below the objects .
Let u is the velocity of 1 kg block before collision and v is the velocity of 5 kg block after collision .
Applying conservation of linear momentum .
`1xx u +5 xx 0 =- xx2 +5v`
`rArr " " 5v - u = 2 `
Coefficient of elasticity (e ) is defined as follows :
` -e = (v_(2)-v_(1))/(u_(2)-u_(1))`
Coefficient of elasticity for elastic collision is 1 .
` rArr " " -1 = (v-(-2))/(0-u)`
`rArr " " v+2 = u " " ...(ii) ` .
Substituting from equation (ii) in equation (i) we get the following :
`5v - v - 2 = 2 rArr 4v rArr v = 1 ` m/s .
And substituting above result is equal to momentum of 1 kg block before collision .
Total momentum of system ` = 1xx u = 1xx3 = 3 " kg "ms^(-1)` .
Momentum of 5 kg block after collision `= 5 xx v = 5 xx1 = 5 ` kgm/s .
Velocity of the centre of mass before collision
`V_(cm) = (m_(1)u_(1)+m_(2)u_(2))/(m_(1)+m_(2)) = (1xx3+5xx0)/(1+5) = 3/6 = 0.5 `m/s
Kinetic energy of centre of mass
`KE =1/2(m_(1)+m_(2))V_(cm)^(2) =1/2 (1+5)(0.5)^(2) =0.75 J `
Kinetic energy of the system before collision
`KE =1/2m_(1)u_(1)^(2)+1/2m_(2)u_(2)^(2) =1/2 (1)(3)^(2)+1/5 (5)(0)^(2) = 4.5 J `