Question

A body of mass 3m at rest explodes into two pieces of mass m and 2m.What is the ratio of the kinetic energy of m to that of 2m?

Answer 1

Answer:

The ratio of the kinetic energy of m to that of 2m is 2:1.

Explanation:

Given that,

A body of mass 3m at rest explodes into two pieces of mass m and 2m and we are required to find the ratio of kinetic energies of both the masses.

The initial momentum of the system before the explosion is

$3m\times0=0$

The final momentum of the system after exploding into two pieces is

$mv_{1}+2mv_{2}$

From the Law of conservation of linear momentum, The initial and final momentum are equal.

Therefore,

$mv_{1}+2mv_{2}=0= > v_{1}=-2v_{2}-- > (1)$

Now the ratio of kinetic energies is given as

$\frac{K.E_{m}}{K.E_{2m}} =\frac{\frac{1}{2}mv_{1}^2}{\frac{1}{2}(2m)v_{2}^2}\\ =\frac{v_{1}^2}{2v_{2}^2}$

Substituting equation 1 in the above expression we get

$\frac{K.E_{m}}{K.E_{2m}} =\frac{v_{1}^2}{2v_{2}^2}=\frac{(-2v_{2})^2}{2v_{2}^2} \\=2:1$

Therefore,

The ratio of the kinetic energy of m to that of 2m is 2:1.

#SPJ3