Question

If the momenta of two bodies are in the ratio
3:4 and their kinetic energies in the ratio 25:16.
Then the ratio of their masses is​

Answer 1

Given :

• Ratio of moemta of two bodies = 3 : 4

• Ratio of Kinetic energies of two bdies = 25 : 16

To Find :

• The ratio of their masses

Solution :

Let ,

• the momentum of first and second bodiesbe P₁ and P₂
• the masses of the bodies be m₁ and m₂

We are given that the ratio of momenta of the two bodies as 3 : 4 . This mean ,

$\\ : \implies \sf \: \frac{P_1 }{P_2} = \frac{3}{4} \: \: ............(1)$

The relation between Kinetic energy and momentum is given by ,

$\\ {\boxed{\bf{KE = \dfrac{P^2}{2m}}}}$

We are given that the ratio of kinetic energies of the two bodies as 25 : 16 . This mean ,

$\\ : \implies \sf \: \frac{KE_1}{KE_2} = \frac{25}{16}$

$\\ :\implies \sf \:\dfrac{ \big( \frac{{P_1}^{2} }{2m_1} \big)}{ \big( \frac{{P_2} ^{2}}{2m_2 } \big) } = \frac{25}{16}$

$\\ : \implies \sf \: \frac{{P_1}^{2} \times 2m_2}{ {P_2}^{2} \times 2m_1 } = \frac{25}{16}$

$\\ : \implies \sf \: \frac{ {P_1}^{2} \times m_2}{{P_1 }^{2} \times m_1} = \frac{25}{16}$

$\\ : \implies \sf \: \bigg(\frac{P_1 }{P_2} \bigg)^{2} \times \frac{m_2 }{m_1} = \frac{25}{16}$

From eq(1) ,

$\\ : \implies \sf \: \bigg( \frac{3}{4} \bigg)^{2} \times \frac{m_2}{m_1} = \frac{25}{16}$

$\\ : \implies \sf \: \frac{9}{16} \times \frac{m_2}{m_1} = \frac{25}{16}$

$\\ : \implies \sf \: \frac{m_2}{m_1} = \frac{25}{16} \times \frac{16}{9}$

$\\ : \implies \sf \: \frac{m_2}{m_1 } = \frac{25}{9}$

$\\ : \implies{\underline{\boxed{{\pink {\mathfrak{ \frac{m_1}{m_2 } = \frac{9}{25} }}}}}} \: \bigstar$

Hence ,

• The ratio of masses of the given bodies is 9 : 25.