Question

A block of mass m is placed on an inclined surface

of a wedge of mass M and released from rest. All

surfaces are frictionless. At any instant, the

velocity of wedge is v1 and velocity of block with

respect to wedge is v2 as shown, then relation

between v1 and v2 is _​

Answer 1

Answer:

option c is right answer

Explanation:

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Answer 2

Constraint relation. Approach 1

We can observe that the wedge M can only move in horizontal direction towards left, and the block m can slide on inclined surface of M always in contact with the wedge.

∙ Let us defined our x and y axes parallel to the incline and perpendicular to incline, respectively.

∙ We can observe that the displacement of m and M in -x' direction will be same as the block never lose contact with the wedge.

∙ If the wedge moves in the horizontal direction by a distance x', during this time, the block will move x in x' direction.

∙ We can relate these displacements x and X as

x

y

=sinθ⇒y=xsinθ......(i)

Hence, velocity relation can be written as :

vy=Vsinθ........(ii)

and acceleration relation can be written as :

v

x

=Asinθ.....(iii)

Here v

y

and a

y

are the velocity and acceleration of the block, respectively, in the direction perpendicular to inclined surface,

Approach 2 : We consider the motion of the block parallel to incline and perpendicular to inclined surface. Let the components of acceleration of block with respect to ground along these direction are a

x

and a

y

, respectively.

Then we can write a

y

=Asinθ

Sol for wedge:

Nsinθ=MA

For block : considering the block in the direction perpendicular to sloping surface.

mgcosθ−N−ma

y

But a

y

=Asinθ

Hence,

mgcosθ−N=mAsinθ

From (i|) and (ii), we get

A=

M+msin

2

θ

mgsinθcosθ