Question

A scooter of 40kg mass moving with velocity 4m/s collides with another scooter of 60kg mass and moving with velocity 2 m/s. After the collision the two scooters stick to each other the loss in kinetic energy is
A. 392 J
B. 440 J
C. 48 J
D. 110 J

Answer 1

In the situation given to us in the questions the collision will be inelastic since there is a loss of energy, this lost energy could be changing to the shape of objects involved in collision or sound energy etc. In inelastic collisions the kinetic energy of the system is not conserved and there is some loss in Kinetic energy of the system. The momentum is still conserved for inelastic collision. After collision the scooters stick to each other, their common velocity can be found by the formula given below.

Formula used:
$v=\dfrac{{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}}{\left( {{m}_{1}}+{{m}_{2}} \right)}$

Complete step by step answer:
Collision of bodies can be broadly classified into two different categories, elastic and inelastic. In elastic collision both the momentum and the kinetic energy is conserved and there is no net loss of either of these quantities. But in an inelastic collision the kinetic energy of the system is not conserved and there is some loss in Kinetic energy of the system. The momentum is still conserved for inelastic collision.
Coming to the question at hand. The velocities and masses of the two bodies involved in collision are given

& {{m}_{1}}=40kg \\\ & {{u}_{1}}=4m/s \\\ & {{m}_{2}}=60kg \\\ & {{u}_{2}}=2m/s \\\ \end{aligned}$$ After collision, both the bodies stick together and act as a single body. Applying the law of conservation of momentum, the new velocity, v, of this new system would be $$\begin{aligned} & v=\dfrac{{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}}{\left( {{m}_{1}}+{{m}_{2}} \right)} \\\ & \Rightarrow v=\dfrac{40\times 4+60\times 2}{\left( 40+60 \right)}=2.8m/s \\\ \end{aligned}$$ So, the bodies would move together at 2.8m/s The loss of energy would be the difference in the kinetic energies of the system before and after the Collision. Loss of Energy = $$K{{E}_{1}}-K{{E}_{2}}$$ The difference in Kinetic energies would be $$\begin{aligned} & K{{E}_{1}}-K{{E}_{2}}=\dfrac{1}{2}{{m}_{1}}u_{1}^{2}+\dfrac{1}{2}{{m}_{2}}u_{2}^{2}-\dfrac{1}{2}\left( {{m}_{1}}+{{m}_{2}} \right){{v}^{2}} \\\ & \Rightarrow K{{E}_{1}}-K{{E}_{2}}=\dfrac{1}{2}\times 40\times {{4}^{2}}+\dfrac{1}{2}\times 60\times {{2}^{2}}-\dfrac{1}{2}\left( 40+60 \right)\times {{\left( 2.8 \right)}^{2}} \\\ & \Rightarrow K{{E}_{1}}-K{{E}_{2}}=320+120-392 \\\ & \therefore K{{E}_{1}}-K{{E}_{2}}=48J \\\ \end{aligned}$$ As we can see from the above calculation the difference in the kinetic energies of the system before and after the Collision is 48 Joules. Therefore, the loss of kinetic energy would be 48 J. **So, the correct answer is “Option A”.** **Note:** The situation given in the question of bodies sticking together after collision is a situation they can be encountered frequently. The given solution can be a little lengthy and time consuming in examination halls. So, a direct formula has also been derived to calculate the loss of energy directly. $$loss\text{ }in\text{ }KE=\dfrac{1}{2}\times \dfrac{{{m}_{1}}\times {{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}{{({{u}_{1}}-{{u}_{2}})}^{2}}$$. This formula can be used to calculate loss in kinetic energy directly.