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A ball of mass m=1kg falling vertically ...

A ball of mass `m=1kg` falling vertically with a velocity `v_0=2m//s` strikes a wedge of mass `M=2kg` kept on a smooth, horizontal surface as shown in figure. If impulse between ball and wedge during collision is J. Then make two equations which relate J with velocity components of wedge and ball. Also find impulse on wedges from ground during impact.

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The correct Answer is:
A
Given `M=2kg` and `m=1kg`
Let, J be the impulse between ball and wedge during collision and `v_1`, `v_2` and `v_3` be the components of velocity of the wedge and the ball in horizontal and vertical directions respectively.

Applying `"impulse"="change in momentum"`
we get `Jsin30^@=Mv_1=mv_2`
or `J/2=2v_1=v_2`...`(i)`
`J cos 30^@=m(v_3+v_0)`
or `sqrt3/2J=(v_3+2)` ...`(ii)`
So these are two equations relating J and velocity components of wedge and ball. Further, net vertical impulse on wedge should be zero. Therefore, impulse on wedge from ground is `Jsin30^@` or `J/2` in upward direction.
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