Question

An open watertight railway wagon of mass \[5\times {{10}^{3}}\,kg\]coasts at an initial velocity of $1.2\,m/s$without friction on a railway track. Rain falls vertically downwards into the wagon. What change then occurred in the kinetic energy of the wagon, when it has collected ${{10}^{3}}\,kg$ water?
(A) $1200\,J$
(B) $300\,J$
(C) $600\,J$
(D) $900\,J$

Answer 1

Hint : In this question, we have the value of the mass of the wagon and its velocity along with a mass of water. We are required to find the change in kinetic energy of the wagon and in order to find that we will use the formula of momentum and kinetic energy.

Complete step by step answer:
Given:
 Mass of railway wagon $=5\times {{10}^{3}}\,kg$
 Initial velocity $=1.2\,m/s$
 Mass of water $={{10}^{3}}\,kg$
 Change in kinetic energy of the wagon $=\,?$
Since force in the horizontal direction$=0$,
 ${{P}_{i}}={{P}_{f}}\,\,\,.....(1)$
Where,
 ${{P}_{i}}=$ Initial momentum
 ${{P}_{f}}=$ Final momentum
Momentum is defined as the product of a system’s mass multiplied by its velocity,
 $P=m\times v$
Where,
 $P=$Momentum
 $m=$Mass
 $v=$Velocity
Initial momentum $({{P}_{i}})$ will be the sum of the momentum of the wagon and momentum of the water. Whereas, Final momentum will be equal to the sum of the mass of water and wagon multiplied by velocity.
Since the velocity of water is equal to 0,
Therefore,
 $(5\times {{10}^{3}}\times 1.2)+({{10}^{3}}\times 0)=(6\times {{10}^{3}})v$
 $6\times {{10}^{3}}=(6\times {{10}^{3}})v$
 $v=1\,m/s$
To find the change in kinetic energy, we will find the final and initial kinetic energy.
Final kinetic energy,
 ${{(KE)}_{f}}=\dfrac{1}{2}{{m}_{f}}v_{f}^{2}$
Where,
 ${{m}_{f}}=$ Final mass
 ${{v}_{f}}=$ Final velocity
 ${{(KE)}_{f}}=\dfrac{1}{2}(6\times {{10}^{3}}\times {{1}^{2}})$
 ${{(KE)}_{f}}=3000\,J$
Initial kinetic energy,
 ${{(KE)}_{i}}=\dfrac{1}{2}{{m}_{i}}v_{i}^{2}$
Where,
 ${{m}_{i}}=$Initial mass
 ${{v}_{i}}=$Initial velocity
 ${{(KE)}_{i}}=\dfrac{1}{2}(5\times {{10}^{3}}\times {{(1.2)}^{2}})$
 ${{(KE)}_{i}}=3600\,J$
As we know change in kinetic energy is:
 $\Delta KE={{(KE)}_{i}}-{{(KE)}_{f}}$
 $\Delta KE=3600-3000$
 $\Delta KE=600\,J$

So the change in kinetic energy of the wagon is 600 J. Hence option C is correct.

Note: In order to solve this kind of question we should have knowledge about momentum and kinetic energy. The change in the kinetic energy of an object is equal to the net work done on the object. This fact is referred to as the Work-Energy Principle and is often a very useful tool in mechanics problem-solving.