# If momentum of an object is doubled, then what is the change in its kinetic energy?

Sep 5, 2017

Percentage increase in kinetic energy is 300%.

#### Explanation:

Momentum of a body of given mass, is directly proportiional to its velocity. In fact momentum is mass multiplied by velocity.

Kinetic energy of a body of given mass, is directly proportiional to its square of its velocity.

When momentum is doubled, mass remaining constant, this means velcity is doubled.

Hence, kinetic energy becomes ${2}^{2} = 4 \text{ times}$ i.e. say from $100$ units to $400$ units.

and percentage increase in kinetic energy is 400-100=300%.

Sep 5, 2017

when the momentum is doubled, the kinetic energy goes up by a factor of 4 which represents a 300% increase.

#### Explanation:

Let's start by writing down some equations that we know for momentum and kinetic energy:

Kinetic energy is given by

$K . E . = \frac{1}{2} m {v}^{2}$

while the momentum is

$p = m v$

Both of these quantities is dependent on the mass ,$m$, of the object and the velocity $v$. We can safely assume that the mass of the body is not changing and therefore the change in the momentum is due to the velocity changing. We can break the problem down to two timeframes, 1) before the change and 2) after then change - where the momentum is $2 \times$ more than when we start.

We know the relationship of momentum before and after:

${p}_{2} = 2 {p}_{1}$

$\implies m {v}_{2} = 2 m {v}_{1}$

then we can divide through by the mass to get rid of it in the relationship:

${v}_{2} = 2 {v}_{1}$

What we are looking for is the kinetic energy before and after:

$K . E {.}_{1} = \frac{1}{2} m {v}_{1}^{2}$

$K . E {.}_{2} = \frac{1}{2} m {v}_{2}^{2} = \frac{1}{2} m {\left(2 {v}_{1}\right)}^{2} = 2 m {v}_{1}^{2}$

let's factor out a $\frac{1}{2} m {v}_{1}^{2}$ in the expression for $K . E {.}_{2}$ so that we can substitute $K . E {.}_{1}$

$\implies K . E {.}_{2} = 4 \cdot \left(\frac{1}{2} m {v}_{1}^{2}\right) = 4 \cdot K . E {.}_{1}$

Therefore, when the momentum is doubled, the kinetic energy goes up by a factor of 4. This represents a 300% increase in K.E.

The starting K.E. is set to 100% and we end with 400% of the original K.E. representing a 300% increase.