Question

A stationary particle explodes into two particles of masses x and y which moves in opposition direction with velocity v1 and v2 the ratio of kinetic energy is

Answer 1

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Answer 2

As per the question, the particle explodes into two parts.

The masses of two parts are x and y .

The velocities of the two particle are $v_{1}\ and\ v_{2}\ respectively$

As per the law of conservation of linear momentum, we know that linear momentum of an isolated system always remains constant.

Hence, momentum of the system before explosion is equal to the momentum after explosion.

As per the question, the initial momentum of the particle is zero.

The final momentum of the system of the system after collision is

                                              $xv_{1} +yv_{2}$

As initial momentum = final momentum,

                     ⇒$xv_{1} +yv_{2} =0$

                     ⇒$xv_{1} =\ -yv_{2}$

                     ⇒$\frac{x}{y} =-\frac{v_{2}} { v_{1}}$

Taking magnitude only, we have-

                    ⇒$\frac{x}{y} =\frac{v_{2}} {v_{1}}$

The kinetic energy of a particle is calculated as $K.E=\frac{1}{2} mv^2$

Here, m is the mass and v is the velocity of the particle.

Hence the ratio of kinetic energy of the particle is -

  $\frac{K.E_{x}} {K.E_{y}}= \frac{1}{2}xv_{1}^2/ \frac{1}{2}y v_{2}^2$

                  $=\frac{v_{2}} {v_{1}}\times \frac{v_{1}^2} { v_{2}^2}$

                  $=\frac{v_{1}} {v_{2}}$    [ans]

    This is in terms of velocities. If we want to calculate the ratio in terms of masses, the ratio will be $\frac{y}{x}$