A block of mass `m` is placed on a triangular block of mass `M(M = 2m)` , as shown. All surfaces are smooth. Calculate the velocity of triangular block when the smaller block reaches at bottom.

Answer

A

Explanation

When the block reaches to the bottom of the wedge, their velocities are as shown in figure. Here v is the absolute veloctiy (with respect to ground), but `v_r` is the relative velocity (relative to wedge).

Their absolute velocity components are as shown below:

`v=sqrt((v_rsintheta)^2+(v_rcostheta-v)^2)`
Linear momentum in horizontal direction is conserved.
`:. p_i=p_f`
or `0=m(v_rcostheta-v)-Mv` ...(i)
Mechanical energy is also conserved.
`E_i=E_r`
or `mgh=1/2Mv^2+1/2mv^2`
or `mgh=1/2Mv^2+1/2m[(v, sintheta)^2+(v_rcostheta-v)^2]` ...(ii)
We have two equations and two unknowns v and `v_r`. Solving these equations, we can find the value of v.
Exercise: Solve these two equations to find the value of v.