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Question

A small disc of mass m is placed on a body of mass M at rest. If the disc is set in horizontal motion with velocity v as shown,


(Assume that disc breaks from surface of M and all surfaces are smooth)
Find the height to which disc rises after breaking at point A, from initial level.

A
Mv22g(M+m)
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B
Mv22g(m)
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C
mv22g(m+M)
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D
mv22gM
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Solution

The correct option is A Mv22g(M+m)
When the disk leaves the surface of M, its velocity towards right (along x axis) equals the velocity of the body M.
Let it be vx
Let vy be the disk velocity in vertical direction (along y axis) at the same moment (disc leaving surface).
There is no net external force in horizontal direction, so linear momentum will be conserved for the system
mv=(M+m)vx
vx=mvM+m...(i)
Applying law of conservation of energy for the system,
12mv2=12(M+m)v2x+12mv2y+mgh1....(ii)
Substituting the value of vx from equation (i) in equation (ii)
Where h1 is the height of break off point from initial level
v2y=v2mv2M+m2gh1
Let h2 be the height raised after the break off point, then
v2y=2gh2....(iii)
From equation (ii) and (iii)
h1+h2=Mv22g(M+m)

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