# Question e8598

Oct 1, 2017

41.4%

#### Explanation:

The relation between kinetic energy $K$ and momentum $p$ of an object is
$p = \sqrt{2 m K}$
where m is the mass of object

First let momentum is ${p}_{1} = p = \sqrt{2 m K}$
and Kinetic Energy ${K}_{1} = K$

Now Kinetic Energy of Body is increased by 100%.
Increase in Kinetic energy $= K \times \frac{100}{100} = K$

Now the total Kinetic Energy is ${K}_{2} = K + K = 2 K$
Now Momentum ${p}_{2} = \sqrt{2 m {K}_{2}}$
${p}_{2} = \sqrt{2 m \left(2 K\right)} = \sqrt{2 \left(2 m K\right)} = \sqrt{2} \sqrt{2 m K}$
But $\sqrt{2 m K} = {p}_{1}$
${p}_{2} = \sqrt{2} {p}_{1} = \sqrt{2} p$

Increase in Momentum = ${p}_{2} - {p}_{1} = \sqrt{2} p - p = \left(\sqrt{2} - 1\right) p$
$= 0.414 p$

Percent increase in momentum =0.414p/pxx100%=41.4%#

Oct 2, 2017

A 100% increase means that the Kinetic Energy is doubled. The final Kinetic Energy (${K}_{f}$) will be twice the initial Kinetic Energy (${K}_{i}$).

${K}_{f} = 2 \cdot {K}_{i}$

Assuming that the mass of the object is unchanged, we can determine the momentum ($p$) in each case:

${P}_{i} = \sqrt{2 m \cdot {K}_{i}}$
${P}_{f} = \sqrt{2 m \cdot {K}_{f}} = \sqrt{2 m \cdot 2 {K}_{i}}$

The ratio between the final and initial momentum will give us the fractional increase in momentum.
${P}_{f} / {P}_{i} = \frac{\sqrt{2 m \cdot 2 {K}_{i}}}{\sqrt{2 m \cdot {K}_{i}}} = \sqrt{2} \cong 1.414$

The final momentum will be equal to 1.414 times the initial momentum. In terms of the initial problem statement, this is an increase of about 41.4%.

Notice that if you doubled the mass of the object to increase the Kinetic Energy by a factor of two, the momentum would also double. One assumption you must make with this problem statement is that the mass of the object is unchanged as the kinetic energy is increased. And that's probably the way this question was intended.