Question

The momentum of a body is doubled.

What will be the percentage increase in its Kinetic energy?

Answer 1

Kinetic energy and momentum are related as
K = p²/(2m)

For an object of mass m
K ∝ p²
K₂/K₁ = (p₂/p₁)² ....(1)

Let’s say initial kinetic energy and momentum be 100% each
Then equation (1) will become
K₂/100% = (200% / 100%)²
K₂/100% = 4
K₂ = 400%

Change in kinetic energy = K₂ - K₁
= 400% - 100%
= 300%

∴ Kinetic energy is increased by 300%

Answer 2

Let’s go!

Momentum is given by…

[math]P=mv[/math]

Where [math]P[/math] is momentum, [math]m[/math] is mass, and [math]v[/math] is velocity (this should technically be a vector equation, but I’m going to ignore that. The idea is actually mostly the same either way).

Kinetic energy is given by…

[math]KE=\frac{1}{2}mv^2[/math]

Let’s pick some arbitrary momentum…

[math]P_1=m_1v_1[/math]

Now let’s pick a second one.

[math]P_2=m_2v_2[/math]

We have already decided that [math]P_2=2P_1[/math], and let’s also assume mass is constant, [math]m_2=m_1[/math] so we make those substitutions.

[math]2P_1=m_1v_2[/math]

Now we further substitute for [math]P_1[/math]

[math]2(m_1v_1)=m_1v_2\rightarrow v_2=2v_1[/math]

Cool, so we know that [math]v_2=2v_1[/math]

Now we can look at the Kinetic Energy.

To find the change in something as a percentage of the first value, we write

[math](\frac{a_2}{a_1}-1)*100\%[/math]

So for Kinetic Energy…

[math](\frac{KE_2}{KE_1}-1)*100\%[/math]

Make the appropriate substitutions…

[math](\frac{\frac{1}{2}m_2{v_2}^2}{\frac{1}{2}m_1{v_1}^2}-1)*100\%[/math]

Make more appropriate substitutions…

[math](\frac{\frac{1}{2}m_1(2v_1)^2}{\frac{1}{2}m_1(v_1)^2}-1)*100\%=300\%[/math]

There you go!