A mass A released from the top of a frictionless incline with angle $\theta$ of length 18 m reaches the bottom in 3 seconds. Just when A is released, a second mass B is projected upwards from the bottom along the same incline with some velocity. This mass B travels a certain distance up the incline, stops there momentarily and returns back such that it reaches the bottom along with the mass A, but the two masses do not collide. What is the initial velocity of B ?

6 m/s

For mass A [descending], using : $S = ut + \frac{1}{2}at^2$

$18 = 0 + \frac{1}{2}\times g \sin {\theta} \times 9 $

$sin\theta = \frac{4}{g}$  ... (1)

For mass B, [ascending], using : $v = u + at$

$0 = u - g \sin {\theta} t_1 \text{ ... [ As : v = 0]}$

$u = g \sin {\theta} t_1$

$t_1 = \frac{u}{gsin\theta}$ ... (2)

Let's say mass B moves $x$ distance up the incline. Using : $v^2 = u^2 - 2g \sin {\theta}$

$0 = u^2 - 2 g \sin {\theta}x$

$u^2= 2gsin\theta x $ ... (3)

For the descend of B, using : $S = ut + \frac{1}{2}\times at^2$

$x = ut_2 + \frac{1}{2}\times g\times \sin {\theta}\times t_2^2 $

$\Rightarrow x = \frac{1}{2}\times g\times \sin {\theta}t_2^2$

$t_2^2 = \frac{2x}{g \sin {\theta}} \text{ ... (4)}$

$t_1 + t_2 = 3 \text{ ... [given]}$

From (2) and (4), we get :

$\frac{u}{g\times \sin {\theta}} + \sqrt{\frac{2x}{g \sin {\theta}}} = 3$

Using the value of $x$ from (3), we get :

$\frac{u}{g\times \sin {\theta}} + \frac{u}{g\times \sin {\theta}} = 3 $

$\Rightarrow \frac{2u}{g \sin {\theta}} = 3$

Using the value of $\sin {\theta}$ from (1), we get :

$\frac{2u}{g}\times \frac{g}{4} = 3$

$\Rightarrow u = 6 m/s$