Question #2d4b3

Jul 1, 2017

The kinetic energy will be halved.

Explanation:

Kinetic energy is given by:

${E}_{k} = \frac{1}{2} \cdot m \cdot {v}^{2}$

It is proportional to the object's mass:

$E \propto m$

If we double the mass, we double the energy:

${E}_{1} = \frac{1}{2} \cdot 1 \cdot {v}^{2} = \frac{1}{2} {v}^{2}$

${E}_{2} = \frac{1}{2} \cdot 2 \cdot {v}^{2} = 1 {v}^{2} = \textcolor{b l u e}{2 {E}_{1}}$

Kinetic energy is also proportional to the square of the velocity:

$E \propto {v}^{2}$

This means that if we halve the velocity, we quarter the energy:

${E}_{1} = \frac{1}{2} \cdot m \cdot {1}^{2} = \frac{1}{2} \cdot m$

${E}_{2} = \frac{1}{2} \cdot m \cdot {\left(\frac{1}{2}\right)}^{2} = \frac{1}{2} \cdot m \cdot \left(\frac{1}{4}\right) = \frac{1}{8} \cdot m = \textcolor{b l u e}{\frac{1}{4} \cdot {E}_{1}}$

If we combine a doubled mass and halved velocity, our initial energy will be doubled and then quartered. This results in an overall halving of the energy:

${E}_{2} = {E}_{1} \cdot 2 \cdot \frac{1}{4} = \frac{1}{2} {E}_{1}$