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Question

A ball of mass m is suspended by a massless string of length l=10 m from a fixed point as shown in the figure below. A ball of mass 2m strikes at an angle θ=45 from the horizontal and sticks to it. If the system deflects by angle ϕ=π2, then the velocity of the system just after the collision and the velocity of 2m particle before collision are, respectively (in m/s)


A
102,30
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B
10,302
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C
10,30
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D
102,302
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Solution

The correct option is A 102,30
Assume the velocity of mass 2m initially is u and just after the collision, common velocity of the system is v.

Immediately after collsion, vertical velocity of the system (m+2m) will be zero due to the string.
By momentum conservation along horizontal direction :-
(2mcos45)u=(m+2m)v
2u×12=3v
or u=32v ...(1)

By energy conservation after the collision,
Initial Kinetic Energy = Final Potential Energy
12(3m)v2=3mgl
(since system deflects by angle π2, final height achieved is l)

v=2gl=2×10×10
v=102 m/s

Velocity of system just after collision
v=102 m/s in horizontal direction

and velocity of 2m mass particle initially, will be
u=32×v=32×102=30 m/s
(from (1))

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