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Question: A particle of mass m is moving with speed 2v and collides with a mass 2m moving with speed v in the ...

A particle of mass m is moving with speed 2v and collides with a mass 2m moving with speed v in the same direction. After collision, the first mass is stopped completely while the second one splits into two particles each of mass m which moves at an angle of 45 degree with respect to the original direction. The speed of the moving particle will be?
A. v22\dfrac{v}{{2\sqrt 2 }}
B. 22v2\sqrt 2 v
C. 2v\sqrt 2 v
D. v2\dfrac{v}{{\sqrt 2 }}

Explanation

Solution

Hint- For solving this numerical we will mainly use the momentum conservation which is nothing but simply a statement of Newton's third law of motion. During a collision the forces on the colliding bodies are always equal and opposite at each instant.

Step-By-Step answer:
As we know the forces cannot be anything but equal and opposite at each instant during collision. Hence the impulses (force multiplied by time) on each body are equal and opposite at each instant and also for the entire duration of the collision.
Impulses of the colliding bodies are nothing but changes in momentum of colliding bodies. Hence changes in momentum are always equal and opposite for colliding bodies. If the momentum of one body increases then the momentum of the other must decrease by the same magnitude. Therefore the momentum is always conserved.
By applying the momentum conservation in horizontal direction
Momentum of the body before collision
=2mv+2mv2mv + 2mv


mvcos450+mvcos450mv'\cos {45^0} + mv'\cos {45^0}
Momentum of the body after collision

Since the momentum of the body is conserved
Momentum before collision = momentum after collision
2mv+2mv2mv + 2mv = mvcos450+mvcos450mv'\cos {45^0} + mv'\cos {45^0}
4v=2v v=22v  4v = \sqrt 2 v' \\\ v' = 2\sqrt 2 v \\\
Hence, the correct option is “B”.

Note- If the two bodies collide in a way that some energy changes from kinetic to something else or if the deformation of the bodies takes place in a way that they cannot recover fully then energy is not conserved. Also remember we can apply momentum conservation only when the net external force on the body is zero.