Question

A block of mass m is placed on a triangular block of mass M, which in turn is placed on a horizontal surface as shown in figure (9-E21). Assuming frictionless surfaces find the velocity of the triangular block when the smaller block reaches the bottom end.

Answer 1

ANSWER::

V₂ = √[2m²g²h cos²α / (M + m)(M + m sin²α)]

Explanation::

Block of mass m will slide down the inclined plane of mass M with an acceleration a₁ g sin α relative to the inclined plane.

Horizontal component of a₁ = g sin α cos α , for which block M will accelerate towards left.

Let acceleration of block M be a₂

According to the concept of centre of mass , in horizontal direction external force is

0ma₁ = (M + m) a₂a₂ = (ma₁) / (M + m)

a₂ = (mg sin α cos α) / (M + m)            [ Equation 1]

Now , the absolute or resultant acceleration of m on block M along direction of incline will be,

a = g sin α - a₂ cos α

a = g sin α - (mg sin α cos² α) / (M + m)

a = g sin α [ 1 - (mcos² α) / (M + m) ]

a = g sin α [ (M + m - m cos²α) / (M +m)]

a = g sin α [ (M + m sin²α) / (M +m)]            [ Equation 2]

Let time taken by block m to reach the bottom end be t .

S = ut + (1/2) at²h / sinα = (1/2)at²t = √( 2 / a sin α)

Now , the velocity of bigger block (V₂) after time t will be ,

V₂ = u + a₂t = [(mg sin α cos α) / (M + m)] √(2h / a sin α)

V₂= √[2m²g²h sin²α cos²α) / {(M + m)² a sin α}]

Now , subtracting value of a from equation 2 we get ,

V₂ = √[ 2m²g²h sin²α cos²α (M + m) / (M + m)² sin α g sin α (M + m sin²α)]

V₂ = √[2m²g²h cos²α / (M + m)(M + m sin²α)]

Hope it helps!