Question

A balloon of mass M is stationary in the air. It has a ladder on which a man of mass ‘m’ is standing. If the man starts climbing up the ladder with a velocity ‘v’ relative to the ladder, the velocity of balloon is
A. $\dfrac{mv}{M}$ upwards
B. $\dfrac{mv}{M+m}$ downwards

Answer 1

Hint:Momentum is defined as the product of mass and velocity of the object. It depends on the direction in which the object is moving. So it is a vector quantity.

Complete step by step solution:
If two objects are colliding with each other, then the momentum of the object before collision and the momentum after collision, remains constant or conserved.
Given that the mass of balloon is = M
Given that the Mass of man is = m
Given that the Velocity with which the man is climbing the ladder is = v
Suppose the velocity of the balloon is = v’

Using the law of conservation of momentum, we can write that;
Total momentum of system before collision = Total momentum of system after collision
$(M+m)v’=mv$
\[v' = \dfrac{{mv}}{{M + m}}\] in a downward direction.
If the man is going upwards, then the velocity of balloon is,
\[v' = \dfrac{{mv}}{{M + m}}\]

Therefore, Option b is the correct answer.

Note: The momentum of an object is represented by the symbol ‘p’ and is written by the formula p=mv; where ‘m’ is the mass and ‘v’ is the velocity. The more is the mass and the more the velocity, the more is the momentum of the object. The momentum of an object changes, when force is applied on the object. If the total force acting on the objects in a system is zero, then the momentum of the system will be conserved or remain constant. Due to any reason, if the value of momentum changes, then it is known as impulse.