A mass m1 with initial speed v0 in the positive x−direction collides with a mass m2=2m1 which is initially at rest at the origin, as shown in figure. After the collision m1 moves off with speed v1=v02 in the negative y-direction, and m2 moves off with speed v2 at angle θ. (v0=5ms−1) (v0=3ms−1)Find the magnitude of velocity of the centre of mass in ms−1 after the collision.

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A mass $m_{1}$ with initial speed $v_{0}$ in the positive $x - d i r e c t i o n$ collides with a mass $m_{2} = 2 m_{1}$ which is initially at rest at the origin, as shown in figure. After the collision $m_{1}$ moves off with speed $v_{1} = \frac{v_{0}}{2}$ in the negative y-direction, and $m_{2}$ moves off with speed $v_{2}$ at angle $θ . (v_{0} = \sqrt{5} m s^{- 1})$ $(v_{0} = 3 m s^{- 1})$
Find the magnitude of velocity of the centre of mass in $m s^{- 1}$ after the collision.

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Detailed Solution

(a) Velocity of centre of will be same before collision and after collision $v_{e m} = \frac{m_{1} v_{0} + m_{2} \times 0}{m_{1} + m_{2}} = \frac{m_{1} v_{1}}{(m_{1} + 2 m_{1})} = \frac{v_{0}}{3}$

The velocities of centre of mass will be remain unchanged after collision.

(b) Conserving linear momentum

In x-direction: $m_{1} v_{0} = m_{2} v_{2} \cos θ o r v_{2} \cos θ = \frac{v_{0}}{2}$

In y-direction: $0 = m_{2} v_{2} \sin θ - m_{1} \frac{v_{0}}{2} o r v_{2} \sin θ = \frac{v_{0}}{4}$

Squaring and adding (i) and (ii), we get, ${v^{2}}_{2} = {(\frac{v_{0}}{2})}^{2} + {(\frac{v_{0}}{4})}^{2} \Rightarrow v_{2} = \frac{\sqrt{5}}{4} v_{0}$