Question

Two identical 4kg blocks are moving with the same velocity \[1m/s\] towards each other along a smooth horizontal surface. When they collide they stick together and come to rest. The work done by external and internal forces will be respectively
A) \[0,4J\]
B) \[4J,9J\]
C) \[4J,0\]
D) \[0,9J\]

Answer 1

Hint: Recall the principle of conservation of energy, that energy can never be created or destroyed. It occurs whenever two bodies interact with each other. This means that the sum of kinetic and potential energy in a system always remains the same. This principle will be followed here.

Complete step by step answer:
Step I:
The law of conservation of energy states that energy can be transferred from one form to another. Since the bodies are colliding with each other and then coming to rest, their energy also changes but .total energy remains constant.
Step II:
Let \[{m_1}\& {m_2}\] be their masses each \[ = 4kg\]
Let \[{u_1}\& {u_2}\] be their initial velocities before collision \[ = 1m/s\]
Since after collision they come to rest, so their final velocities are zero\[v = 0\]
Step III:
Initial kinetic energy of the system is \[K{E_1} = \dfrac{1}{2}mu_1^2 + \dfrac{1}{2}mu_2^2\]
\[K{E_1} = \dfrac{1}{2} \times 4 \times {(1)^2} + \dfrac{1}{2} \times 4 \times {(1)^2}\]
\[K{E_1} = 2 + 2\]
\[K{E_1} = 4J\]
Step III:
Final kinetic energy \[K{E_2} = \dfrac{1}{2} \times 4 \times {(0)^2} + \dfrac{1}{2} \times 4 \times {(0)^2}\]
\[K{E_2} = 0\]
Step IV:
According to the work-energy theorem, the net work done by the total internal and external forces on an object is equal to the change in kinetic energy.
Step V:
Net work is done = Change in kinetic energy
Change in kinetic energy is given by \[\Delta KE = K{E_2} - K{E_1}\]---(i)
Step VI:
Substituting values of \[K{E_1}\& K{E_2}\] in equation (i),
\[\Delta KE = 0 - 4\]
\[\Delta KE = - 4J\]
The negative sign shows that both the bodies are moving in the opposite direction.

$\therefore$ The work done by the external and internal forces is \[0,4J\]. So, option (A) is correct.

Note:
Here work-energy theorem is important as when two objects move, work is said to be done. When work is done, then its kinetic energy increases. This increase in kinetic energy is known as ‘energy of motion’. But when the objects come to rest then their kinetic energy decreases. From here arises the relation of work-energy theorem and hence it is important.