A body of mass 'm' moving with a velocity 'v' in the x-direction collides with a

Question

A body of mass 'm' moving with a velocity 'v' in the x-direction collides with a body of mass 'M' moving with a velocity 'V' in the y-direction. They stick together during a collision. Then

Options:

A . The magnitude of the momentum of the composite body is √(mv)2+(MV)2

B . The composite body moves in a direction making a angle θ=tan−1(MVmv) with the x-axis.

C . The loss of kinetic energy as a result of collision is 12Mm(M+m)(V2+v2)

D . All the above choices are correct.

Answer: Option D
:
D
Refer to Figure Here p = mv and P = MV.

A Body Of Mass 'm' Moving With A Velocity 'v' In The X-direc...

The resultant of p and P is

p_{r} = \sqrt{p^{2} + P^{2}} = \sqrt{(m v)^{2} + (M V)^{2}}

Which is choice (a), the angle which the resultant momentum

p_{r}

subtends with the x-axis is given by

t a n θ = \frac{P}{p} = \frac{M V}{m v}

,which is choice (b).
Loss of KE =

(\frac{1}{2} m v^{2} + \frac{1}{2} M V^{2}) - [\frac{1}{2} \frac{m^{2} v^{2} + M^{2} V^{2}}{(M + m)}]

\frac{1}{2} \frac{M m}{(M + m)} (V^{2} + v^{2})

,which is choice (c).
Hence the correct choice is (d).

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