Question

A railway wagon of mass $2000kg$ moving with velocity of $18km/h$ collides head on this stationary wagon of mass $3000kg,$ if the two wagon more together after collision,
calculate (i) their common velocity and (ii) loss of Kinetic energy due the collision.

Answer 1

Hint: Use the law of conservation of momentum to solve this question. It states, for a collision occurring between object one and object two, in an isolated system, the total momentum of the two objects before the collision and after the collision is equal.

Formula used:
Law of conservation of momentum
$({m_1} + {m_2})V = {m_1}{v_1} + {m_2}{v_2}$

Complete step by step answer:
According to the law of conservation of momentum.
for a collision occurring between object one and object two, in an isolated system, the total momentum of the two objects before the collision and after the collision is equal.
Mathematically, it is written as
$({m_1} + {m_2})V = {m_1}{v_1} + {m_2}{v_2}$ . .. (1)
Where,
${m_1}$ is the mass of object one
${m_2}$ is the mass of object two
${v_1}$ is the velocity of object one
${v_2}$ is the velocity of object two
Let, the railway wagon be the object one and the stationary wagon be the object two.
Then we have,
${m_1} = m = 2000$
${m_2} = 3000$
${v_1} = 18km/h = 5m/s$
${v_2} = 0$ as the other wagon is stationary.
Therefore, equation (1) gives
$(2000 + 3000)V = 2000 \times 5 + 3000 \times 0$
$ \Rightarrow 5000V = 2000 \times 5$
$ \Rightarrow V = \dfrac{{10000}}{{5000}}$
$ \Rightarrow V = 2m/s$
Initial kinetic energy is given by
${K_1} = \dfrac{1}{2}{m_1}{v_1}^2$
$ = \dfrac{1}{2}2000 \times {5^2}$
$ \Rightarrow {K_1} = 25000J$
The final kinetic energy is given by
${K_2} = \dfrac{1}{2}({m_1} + {m_2}){V^2}$
$ = \dfrac{1}{2}5000 \times {2^2}$
$ \Rightarrow {K_2} = 10000J$
Therefore, the loss of kinetic energy due to the collision.
$\Delta K = {K_1} - {K_2}$
$ = 25000J - 10000J$
$ = 15000J$
$\Delta K = 15J$
Therefore,
(i) There common velocity is $2m/s$ and
(ii) The loss of kinetic energy due to the collision is $15J$

Note:This was a simple question of substituting values on the formula. But you could solve it if you know that the momentum is conserved. So it is important to know the laws and definitions. Also, you should not forget to convert velocity into SI units. As all the rest of the quantities were in SI units.