(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A block, mass m, placed on an wedge, mass M, which is placed on a horizontal surface. All surfaces are frictionless. We have to find the velocity of the wedge when the block reaches the ground. Height of the wedge is h, and the angle of wedge is β. The diagram is in the attachment.

2. Relevant equations

Conservation of linear momentum in the horizontal direction, relative velocity and general kinematics equations. Some variables:

N = normal reaction between block and wedge

3. The attempt at a solution

(1)The block and wedge can be treated as particles.

Considering the block+wedge as a system, wrt ground, no external force is acting in the horizontal direction. Thus, applying conservation of momentum in horizontal direction, we get

m*0 + M*0 = m*(velocity of block at bottom of wedge wrt ground in horizontal direction) + M*(Required velocity) .... (1) [Since the wedge is constrained to move along the horizontal]

(2)Now the acceleration of the wedge wrt ground is Nsinβ.

wrt the ground, the net force on the block along the incline is gsinβ at all times and N = mgcosβ at all times.

Applying the concept of 'pseudo' force in non-inertial reference frames, acceleration of block wrt wedge along the incline at all times is gsinβ(1+((mcos^{2}β)/M)

Since this acceleration is constant at all times, we can apply the kinematics equation V^{2}-U^{2}=2as, wrt the wedge from the beginning to the time the block reaches the bottom. This gives us the velocity of the block along the incline wrt the wedge at the bottom of the incline, which is

[2gh(1+((mcos^{2}β)/M)]^{1/2}. [let's call this k]

(3) Since the block is constrained to move along the incline, this is the actual velocity of the block along the incline wrt the wedge at the bottom of the incline. Now taking it's component in the horizontal direction, we get kcosβ. Now using the concept of relative velocity, velocity of block wrt ground when it is at the bottom in the horizontal direction is kcosβ + Required velocity.

(4) Substituting this in the equation (1),

The required velocity comes out as (m/M+m) * kcosβ, which is

[[2ghmcos^{2}β(M+mcos^{2}β)]/[M(M+m)^{2}]]^{1/2}

However, the answer is given as

[[2ghmcos^{2}β]/[(M+m)(M+msin^{2}β]]^{1/2}

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# Homework Help: Conservation of linear momentum problem

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