Question

A bomb at rest explodes into two parts of masses m1 and m2 , if the momentum of two part be p1 and p2 then find the kinetic energy

Answer 1

hope this will help u let mass of part 1 is m1 and velocity is v1 and so on

Answer 2

The correct answer to the question is $\frac{p_{1}^2} {2m_{1}} +\frac{p_{2}^2} {2m_{2}}$      

CALCULATION:

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

The masses of two parts are given as $m_{1}\ and\ m_{2}\ respectively$

The momentum of two parts are given as $p_{1}\ and\ p_{2}\ respectively$.

We are asked to calculate the kinetic energy.

The momentum of  a body is defined as the product of mass with velocity.

Mathematically momentum p = mv.

The kinetic energy of  a body moving with velocity v is calculated as

                                      $K.E=\ \frac{1}{2}mv^2$

                                             $=\ \frac{1}{2} \frac{m^2v^2}{m}$

                                             $=\ \frac{p^2}{2m}$

Hence, the kinetic energy of the first part $K.E_{1}=\ \frac{P_{1}^2} {2m_{1}}$

The kinetic energy of the second part $K.E_{2}=\ \frac{p_{2}^2} {2m_{2}}$

Hence, the total kinetic energy K.E = $K.E_{1} +K.E_{2}$

                                                           = $\frac{p_{1}^2} {2m_{1}} +\frac{p_{2}^2} {2m_{2}}$         [ans]