Two particles of masses m and 2m moving in opposite direction collides elastically with velocities 2v and v respectively. Find their respective magnitude of velocities after the collision.
A
2v,v
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B
v,2v
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C
v,v
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D
4v,0
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Solution
The correct option is A2v,v Let the final velocities of m and 2m be v1 and v2 respectively.
By conservation of momentum in x−direction, we get:
m(2v)+2m(−v)=m(v1)+2m(v2)
0=mv1+2mv2
∴v1+2v2=0(1)
Also the collision is elastic, ⇒speed of separation = speed of approach
∴v2−v1=2v+v
Or, v2−v1=3v(2)
Solving the above two equations, we get
v1=−2v and v2=v
Hence, mass m moves with speed 2v along −ve,x−direction and mass 2m moves with speed v in the +vex−direction.