Question

Two bodies of masses m and 4 m are moving
with equal linear momentum. The ratio of
their kinetic energy is​

Answer 1

Given :

Two bodies of masses m and 4m are moving with equal linear momentum.

To Find :

The ratio of their kinetic energy.

Solution :

❒ First of all we need to derive relation between kinetic energy and linear momentum of body.

Let a body of mass m is moving at a speed of v.

Kinetic energy of the body :

➠ k = 1/2 mv² .......... (I)

Linear momentum of the body :

➠ p = mv .......... (II)

On squaring both sides, we get

➠ p² = m²v²

From the first equation, m²v² = 2mk

➠ p² = 2mk

➠ k = p² / 2m

Let's come to the question :D

$\sf:\implies\:\dfrac{k_1}{k_2}=\dfrac{{p_1}^2/2m_1}{{p_2}^2/2m_2}$

Given that, p₁ = p₂

$\sf:\implies\:\dfrac{k_1}{k_2}=\dfrac{{p}^2/2m_1}{{p}^2/2m_2}$

$\sf:\implies\:\dfrac{k_1}{k_2}=\dfrac{m_2}{m_1}$

$\sf:\implies\:\dfrac{k_1}{k_2}=\dfrac{4m}{m}$

$:\implies\:\underline{\boxed{\bf{\purple{k_1:k_2=4:1}}}}$

Answer 2

Solution :-

As per the given data ,

• Mass of body 1 = m
• Mass of body 2 = 4m

Both the bodies have equal linear momentum .

Let the linear momentum of both the bodies be P

As per the formula  ,

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

For Body I

$\leadsto KE = \dfrac{P^{2} }{2m }... (1)$

For Body II

$\leadsto KE' = \dfrac{P^{2} }{2 \times 4 m }\\ \\ \leadsto KE ' = \dfrac{P^{2} }{8m }... (2)$

Dividing (1) and (2) we get ,

$\leadsto \dfrac{KE}{KE'} = \dfrac{P^{2} }{2m } \times \dfrac{8m }{P^{2} } \\ \\ \leadsto \boxed{\dfrac{KE}{KE'} = \dfrac{4 }{1}}$

Kinetic energy of  the masses are in the ratio 4:1