Question

A block of mass is projected on the smooth ground with speed Vo towards a smooth incline of mass at rest. If block always remains in contact with wedge when block again returns to the ground is:

Answer 1

Hint: The law of conservation of linear momentum states when no external forces are acting on the system of two colliding objects then the vector sum of the linear momentum of each body remains constant.
For individual bodies, the momentum might increase or decrease according to the situation, but the momentum of the system will always be conserved, as long as there is no external net force acting on it.
The requirement for momentum conservation is that during the interaction the mass of the system must remain constant and the net external force on the system should be zero.
Momentum is a vector quantity, and therefore we need to use vector addition when summing together the momentum of multiple bodies which make up a system.

Formula used:
$\mathop m\nolimits_1 \mathop u\nolimits_1 + \mathop m\nolimits_2 \mathop u\nolimits_2 = \mathop m\nolimits_1 \mathop v\nolimits_1 + \mathop m\nolimits_2 \mathop v\nolimits_2 $

Complete step-by-step solution:
Mass of the block = \[m\]
The velocity of block = $\mathop v\nolimits_0 $
So, initial momentum of block= mass of block multiplied by the velocity of the block
                                                      = $\mathop {m \times v}\nolimits_0 $
The block moves towards the wedge and then it remains in contact with the wedge.
When the block is in contact with the wedge total mass of the system = (m+2m) =3m.
Let the overall system move with velocity V.
Again when the block returns to the ground then the velocity of the block is in the opposite direction.
Hence momentum, in that case, = $ - \mathop {m \times v}\nolimits_0 $ [negative sign indicates the opposite direction of motion of body]
Hence applying conservation of linear momentum.
$m{v_0} = 3mv - m{v_0}$
$ \Rightarrow 2m{v_0} = 3mv$
$ \Rightarrow v = \dfrac{{2{v_0}}}{v}$

So, the velocity of the wedge after the block returns to the ground is $v = \dfrac{{\mathop {2v}\nolimits_0 }}{3}$.

Note: Momentum conservation is a direct consequence of Newton’s Third Law. The total linear momentum of a system equals the product of the velocity of the center of mass and total mass of the system. It should always be ensured that in a system the initial momentum is always equal to the final momentum. Application of conservation of linear momentum can be found in the launching of rockets where the rocket fuel burns and exhausts the gas in a downward direction and because of this the rocket gets pushed in the upward direction. The motorboat pushes the water backward and gets pushed forwards in reaction to conserve momentum. So it follows the same principle.