Two particles A and B of masses 10 kg and 38 kg, respectively, are moving along the same straight line with velocities 15 m/s and 3 m/s, respectively, in the same direction. After elastic collision, the velocities of A and B are `v_A `and `v_B`, respectively, in the direction of initial motion. Then

Options

A. `v_A = 20 m//s, v_B = 8 m/s `
B. `v_A = – 4 m//s, v_B = 8 m//s`
C. `v_A = 16 m/s, v_B = 28 m//s`
D. `v_A = – 5 m//s, v_B = 10 m//s`

Answer

B

Explanation

Using the values given in question and applying momentum conservation we get

`10 xx 15 + 38 xx 3 = 10 xx v_A + 38 xx v_B " ".......(i)`
Now in elastic collision Velocity of approach = Velocity of seperation
We get
`15 – 3 = v_B - v_A" " ....(ii)`
On solving (i) & (ii) equation
`v_A = - 4m//s , v_B = 8 m//s`
Alternate solution
Apply Direct Formula
`v_A = (m_A - m_B)/(m_A + m_B) u_A + (2m_B)/(m_A + m_B) u_B`
`v_B = (2m_A)/(m_A + m_B) u_A + (m_B + m_A)/(m_A + m_B) u_B`

where `u_A = 15m//s , u_B = 3m//s`
on substituting values we get
`v_A = - 4m//s , v_B = 8 m//s`