Question

A stationary particle explodes into two particles of masses x and y, which move in opposite directions with velocity ${{v}_{1}}$ and ${{v}_{2}}$. The ratio of their kinetic energy $\left( {{E}_{1}}:{{E}_{2}} \right)$ is,
$\text{A}.1$
$\text{B}\text{.}\dfrac{x{{v}_{2}}}{y{{v}_{1}}}$
$\text{C}\text{.}\dfrac{x}{y}$
$\text{D}\text{.}\dfrac{x}{y}$

Answer 1

Hint: Use the law of conservation of momentum. It states that the total momentum of an isolated system remains constant; that is, momentum can neither be created nor can be destroyed.

Complete step by step answer:
Velocity of the particle before the explosion is zero.
Since, $\text{Linear momentum}=\text{mass}\times \text{velocity}$
So, the momentum of the particle before the explosion is also zero.
Conservation of momentum gives that total linear momentum of a given system always remains constant. So, we must have the same total momentum before and after the explosion.
Hence, $0=x\times {{v}_{1}}+y\times {{v}_{2}}$
Here, ${{v}_{1}}$is the velocity of particle x and ${{v}_{2}}$ is the velocity of particle y and we considered $v_2$ to be negative since the particles are moving in opposite direction.
$x{{v}_{1}}=y{{v}_{2}}$
Since, we need the ratio of kinetic energies we need to convert the above equation in terms of kinetic energy of the two particles.
Squaring both sides, ${{\left( x{{v}_{1}} \right)}^{2}}={{\left( y{{v}_{2}} \right)}^{2}}$
Now, multiplying both sides by $\dfrac{1}{2}$,
$\dfrac{1}{2}{{\left( x{{v}_{1}} \right)}^{2}}=\dfrac{1}{2}{{\left( y{{v}_{2}} \right)}^{2}}$
$\dfrac{1}{2}x{{v}_{1}}^{2}\times x=\dfrac{1}{2}y{{v}_{2}}^{2}\times y$
${{E}_{1}}\times x={{E}_{2}}\times y$
$\dfrac{{{E}_{1}}}{{{E}_{2}}}=\dfrac{y}{x}$
Where, ${{E}_{1}}$ & ${{E}_{2}}$ are the kinetic energies of the particles x and y respectively.
Hence, the ratio of their kinetic energy $\left( {{E}_{1}}:{{E}_{2}} \right)$ is $\dfrac{y}{x}$
Hence, the answer is option (D)

Additional information:
Law of conservation of momentum, law of conservation of energy & law of conservation of angular momentum can be applied to an isolated system. An isolated system can be defined as a collection of matter which does interact with the universe outside the system.

Note: In this type of question first try to imagine if the conservation laws such as law of conservation of momentum or law of conservation of energy can be used to solve the question.
Conservation laws tell us that certain physical properties of an isolated system remain constant over time.