Question

Two bodies having masses m and 3m resp. are moving such that ratio of their momentum are 1:9. Find the ratio of their kinetic energies ​

Answer 1

GIVEN,

Mass of two bodies are in m and 3m.

Ratio of their momentum is 1:9

Let the common ratio be x,

so,

Momentum of first body = 1x

Momentum of second body = 9x

We know,

$Kinetic \:energy =\dfrac{1}{2} \times m \times v^{2}$

where m is the mass of the body and v is its velocity.

Since,

Momentum p= mv

$v= \dfrac{p}{m}$

On subsituting value of v ,

$Kinetic \:energy =\dfrac{1}{2} \times m \times(\dfrac{p}{m})^{2}$

$Kinetic \:energy =\dfrac{p^{2}}{2m}$

For first body,

$Kinetic \:energy \: k_1 =\dfrac{1^{2}}{2m}$

For second body,

$Kinetic \:energy \: k_2 =\dfrac{9^{2}}{2 \times 3m}$

$Kinetic \:energy \: k_2 =\dfrac{81}{6m}$

On dividing kinetic energy of first body by second,

$\dfrac{K_1}{K_2} =\dfrac{1}{2m} \div \dfrac{81}{6m}\\\\\dfrac{K_1}{K_2} =\dfrac{1}{2m} \times \dfrac{6m}{81}\\\\\dfrac{K_1}{K_2} =\dfrac{6}{162} \\\\\dfrac{K_1}{K_2} =\dfrac{1}{27}$

So the ratio of their kinetic energies will be 1:27.