Question

A body of mass 5 kg is moving with a velocity of 10 m/s.What will be the ratio of its initial and final kinetic energy, if the mass is doubled and its velocity is halved ?

Answer 1

KE initial = 1/2mv²
=1/2×5×10²
=250J

KE final=1/2×2m×(1/2v)²
=2/4×(1/2mv²)
=2/4×Ke initial
=250/2J
=125J

Answer 2

Answer:

$\dfrac{K_1}{K_2}=\dfrac{2}{1}$

Explanation:

It is given that,

Initial mass of the body, m₁ = 5 kg

Initial velocity of the body, v₁ = 10 m/s

Final mass of the body, m₂ = 2m₁ = 10 kg

Final velocity of the body, v₂ = (1/2)v₁ = 5 m/s

Initial kinetic energy of the body is given by :

$K_1=\dfrac{1}{2}\times m_1\times v_1^2$

$K_1=\dfrac{1}{2}\times 5\times (10)^2$..............(1)

Final kinetic energy of the body is given by :

$K_2=\dfrac{1}{2}\times m_2\times v_2^2$

$K_2=\dfrac{1}{2}\times 10\times (5)^2$............(2)

Dividing equation (1) and (2) as :

$\dfrac{K_1}{K_2}=\dfrac{\dfrac{1}{2}\times 5\times (10)^2}{\dfrac{1}{2}\times 10\times (5)^2}$

$\dfrac{K_1}{K_2}=\dfrac{2}{1}$

So, the ratio of its initial and final kinetic energy is 2:1. Hence, this is the required solution.