A small particle of mass `m = 2 kg` moving with constant horizontal velocity `u = 10 m//s` strikes a wedge shaped block of mass `M = 4 kg` placed on smooth horizontal surface on its inclined surface as shown in figure. After collision particle starts moving up the inclined plane. Calculate the velocity of wedge immediately after collision.

Answer

2

Explanation

Let `V_("rel")` be the final velocity of the ball w.r.t. wedge and `V` be the final velocity of the wedge w.r.t. ground. Now, velocity of ball w.r.t. ground

Horizontal component `= V_(x) = V_("rel"). Cos alpha + V`
Vertical component `=V_(y) = V_("rel".) sin alpha`
COM in horizontal direction gives
`m u = m(V_("rel") cos alpha + V) + MV` ........(i)
Since velocity of ball along wedge remains constant
`therefore u cos alpha = V_("rel".) + V coa alpha` ...........(ii)
Solving (i) & (ii) we get
`V = ( m u sin^(2) alpha)/(M+m sin^(2) alpha) = 2m//s`.