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Question: A girl of mass \(50\,kg\) jumps out of a moving boat of mass \(300\,kg\) to the bank with a horizont...

A girl of mass 50kg50\,kg jumps out of a moving boat of mass 300kg300\,kg to the bank with a horizontal velocity of 3ms13\,m{s^{ - 1}}. With what velocity does the boat begin to move backwards?

Explanation

Solution

To answer this question, we first need to understand the concept of the conservation of momentum. Momentum is conserved when a collision occurs in an isolated system. The cumulative sum of momentum in the system's set of objects is the same before and after the collision. A popular physics lab involves dropping a brick onto a moving cart.

Complete step by step answer:
As given in the question
M1{M_1}= 50 Kg (mass of girl)
M2\Rightarrow {M_2}= 300 Kg (mass of boat)
V1=3ms1\Rightarrow {V_1} = 3\,m{s^{ - 1}}.........(velocity of girl)
V2\Rightarrow {V_2} = To find
As per law of conservation of momentum net momentum is equal to zero.

Principle of conservation of momentum: If there is no external force acting on the colliding objects, the theory of conservation of momentum states that if two objects collide, their total momentum before and after the collision will be the same.
M1V1+M2V2=0{M_1}{V_1} + {M_2}{V_2} = 0
Putting the given values
50kg×3ms1+300Kg×V2=050\,kg \times 3m{s^{ - 1}} + 300Kg \times {V_2} = 0
150kgms11300Kg=V2150\,kgm{s^{ - 1}}\dfrac{1}{{300Kg}} = - {V_2}
Hence V2=0.5ms1{V_2} = - 0.5\,m{s^{ - 1}}.....(negative sign represents the direction i.e. backwards)

Hence, the velocity with which the boat begins to move backwards is 0.5ms10.5\,m{s^{ - 1}}.

Note: The net force acting on a system of particles is often zero, or very similar to it. It's similar to energy conservation in that momentum can be transferred from one particle to another within the system. When this occurs, though, the overall momentum must remain constant.