A block of mass 1 kg is moving towards a movable wedge of mass 2 kg as shown in figure. All surfaces are smooth. When the block leaves the wedge from top, its velocity is making an angle `theta =30^(@)` with horizontal.

The value of `v_(0)` in m/s is

Options

A. 4
B. 7
C. 10
D. 9

Answer

B

Explanation

Let v be the horizontal velocity of wedge at topmost point and `v_r` the relative velocity of block with respect to wedge. `v_r` is at `60^@` with horizontal. Absolute velocity of block will be the resultant of `v_r and v` and since it is given at `30^@` with horizontal, `v_r and v` should be equal in magnitude.

Absolute velocity of block `v_b = 2v cos 30^@ = sqrt(3)v`.
Now, applying conservation of linear momentum in horizontal direction and conservation of mechanical energy, we have
`1 xx v_0 = 2 xx v + 1 xx v_b cos 30^@`
or `v_0 = 2v + 3/2 v or " "v_(0) = 7/2 v " "......(1)`
`1/2 xx 1 xx v_0^2 = 1/2 xx 2 xx v^2 + 1/2 xx 1 xx (sqrt(3)v)^2 + 1 xx 10 xx 1.45`
or `v_(0)^2 = 2v^2 + 3v^2 + 29 " or " v_(0)^(2) - 5v^2 = 29 " "....(2)`
Solving Eqs. (1) and (2), we have `v_(0) = 7m//s`.