IIT JEE , AIEEE Forum - Physics , Mechanics

shub

consider a system with a wedge and over it a small block(or any other similar system)

let the block have a velocity u(horizontal component) at any instant.....then the wedge will have a velocity v(say) to conserve the momentum

will we write the equation as-

m(u - v)=(M + m)v ???? or in any other form???

m=mass of block

M=mass of wedge

morningdew

See the initial momentum was zero
And finally the block goes with vel u to the right (or have you taken u to be the relative velocity?), and the block will move to the left with v

So we just write
0 = mu + Mv (taking proper directions)

I guess you are confusing it with cases where suppose a man jumps off a trolley moving with v. THEN, the mass of the man was taken into account in initial momentum because he too was moving with the trolley with same speed.

Here however, the mass m is moving towards right while M moves towards left. By including the m term in RHS [where you write (m+M)v], you would intend that the mass m moves with v as well which it does not.

shub

@morning dew
it's still a bit unclear to me.....
the block is still on the wedge.....so it will also have a velocity v along with the wedge???
and while applying this law...we always take the velocities wrt ground right??????

shub

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