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Question: A ball travelling at \[15{m}/{s}\;\] is returned by a racket in the opposite direction with a veloci...

A ball travelling at 15m/s  15{m}/{s}\; is returned by a racket in the opposite direction with a velocity of 20m/s  20{m}/{s}\;.If the change in KE of the ball thereby is 8.75J8.75J.Find the change in its momentum.

Explanation

Solution

Hint : Firstly, we have to make an expression of the change in kinetic energy by substituting the given values. After that, with the help of this expression, we have to calculate the change in momentum of the ball.

Formula used:
Kinetic Energy, KE=12mv2~KE=\dfrac{1}{2}m{{v}^{2}}
Where, ‘mm’ is the mass of the object in ‘kgkg’ and ‘vv’ is the velocity of that object in ‘m/s  {m}/{s}\;’.

Complete step-by-step solution:
From the question, it is given that :-
Vi=−15m/s  {{V}_{i}}=-15{m}/{s}\;
Vf=20m/s  {{V}_{f}}=20{m}/{s}\;
KE=8.75JKE=8.75J
Therefore, we can conclude that :-
 KE=12mv2~KE=\dfrac{1}{2}m{{v}^{2}}
Now, 12m(Vf2−Vi2)=8.75\dfrac{1}{2}m\left( V_{f}^{2}-V_{i}^{2} \right)=8.75
12m[(Vf−Vi)(Vf+Vi)]=8.75\dfrac{1}{2}m\left[ \left( {{V}_{f}}-{{V}_{i}} \right)\left( {{V}_{f}}+{{V}_{i}} \right) \right]=8.75
m2(Vf−Vi)[20+(−15)]=8.75\dfrac{m}{2}\left( {{V}_{f}}-{{V}_{i}} \right)\left[ 20+(-15) \right]=8.75
m2(Vf−Vi)[5]=8.75\dfrac{m}{2}\left( {{V}_{f}}-{{V}_{i}} \right)\left[ 5 \right]=8.75
m(Vf−Vi)=8.75×25m\left( {{V}_{f}}-{{V}_{i}} \right)=\dfrac{8.75\times 2}{5}
So, here the change in momentum=m(Vf−Vi)=m\left( {{V}_{f}}-{{V}_{i}} \right)
Hence,Δp=3.5kgm/s  \Delta p=3.5{kgm}/{s}\;

Additional information:
Kinetic energy is defined as a form of energy that an object or a particle has by the reason of its motion. If the work that transfers energy is to be done on an object by applying the net force, the object speeds up and thereby gains kinetic energy. Also, Kinetic energy is the property of a moving object or particle and depends not only on its motion but also on its mass. Whereas, Momentum is defined as a product of the mass of a particle and its moving velocity. Also, momentum is a vector quantity which means it has both magnitude and direction. Thus, Isaac Newton’s second law of motion states that the time rate of change of momentum is equal to the force acting on that particle.

Note: Since, the values are not given for the things that are asked in the question. That’s why we have to find the solution by deriving the expression for what is asked by making the expression of the things that are given.