Question

By what percentage does the kinetic energy
increase, if the linear momentum is increased by
50%​

Answer 1

Given :

• Linear Momentum is increased by 50%

To Find :

• Increase in Kinetic Energy

Solution :

As we know that linear momentum is given by,

$\implies \sf{P \: = \: mv} \\ \\ \implies \sf{v \: = \: \dfrac{P}{m} \: \: \: \: \: \: \: ...(1)}$

And, The formula for Kinetic Energy is :

$\implies \sf{K.E \: = \: \dfrac{1}{2} mv^2}$

Now, substitute value of velocity (v) from (1)

$\implies \sf{K.E \: = \: \dfrac{1}{2} m \times \dfrac{P^2}{m^2}} \\ \\ \implies \sf{K.E \: = \: \dfrac{1}{2} \: \times \: \dfrac{P^2}{m}} \\ \\ \implies \sf{K.E \: = \: \dfrac{P^2}{2m}}$

________________________________

If linear Momentum is increased by 50%.

Let new linear Momentum be P' . So, new Momentum will be :

⇒P' = 3/2 P

So, New Kinetic Energy is :

$\implies \sf{K.E' \: = \: \dfrac{(P') ^2}{2m}} \\ \\ \implies \sf{K.E' \: = \: \dfrac{ \bigg( \dfrac{3}{2} P \bigg) ^2}{2m}} \\ \\ \implies \sf{K.E' \: = \: \dfrac{9 P^2}{4 \: \times \: 2m}} \\ \\ \implies \sf{K.E' \: = \: \dfrac{9}{4} K. E}$

Increase in Kinetic Energy is :

$\implies \sf{Increase \: in \: K.E \: = \: \dfrac{K.E' \: - \: K.E}{K.E} \: \times \: 100} \\ \\ \implies \sf{Increase \: in \: K.E \: = \: \dfrac{\dfrac{9}{4} K.E \: - \: K.E}{K.E} \: \times \: 100} \\ \\ \implies \sf{Increase \: in \: K.E \: = \: 125 \: \%}$