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The inclined surfaces of two movable wedges of same mass M are smoothly conjugated with the horizontal plane as shown in the fig. A washer of mass,M slides down the left wedge from a height h. To what maximum height will the washer rise along the right wedge? Neglect friction

Pranav Arora , 8 Years ago
Grade 11
anser 1 Answers
Vikas TU

Last Activity: 8 Years ago

The maximum height it would rise =>
 
let h be the height of wedges.
Now for first wedge momentum conservation and from energy conswrvation,
0 + 0  =  mv1  -  Mv2
we get v2 = mv1/M m/s.
and
mgh = 0.5Mv2^2 + 0.5mv1^2
solving both we get 
v1^2 = 2Mgh/(M+m) …..................(1)
 
For wedge (2)
0.5mv1^2 = 0.5Mv0^2 + mgx
and 
mv1 = Mv0
solving we get,
x = v1^2(M – m)/2gM
substituing from eq. (1)
 we get
 x = [(M-m)/(M+m)]h meter.

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