A bullet traveling at 250 meters per second is brought to rest by an impulse of 5.00 newton-seconds. What is the mass of the bullet?

Jan 17, 2016

The impulse, $5 N s$, is the same as the change of momentum. The final momentum is $0 k g m {s}^{-} 1$, the initial momentum is $250 \cdot m$, so, rearranging, the mass is $\frac{5 k g m {s}^{1}}{250 m {s}^{-} 1} = 0.02 k g \left(= 20 g\right)$.

Explanation:

The change of momentum of an object is equal to the impulse exerted on it.

Impulse is measured in units of newton-seconds $\left(N s\right)$ while momentum is usually expressed as kilogram-metres-per-second $k g m {s}^{-} 1$, however when we realise that the newton is defined as $k g m {s}^{-} 2$, dimensional analysis shows that these two units are the same $\left(N s = k g m {s}^{-} 2 \cdot s = k g m {s}^{-} 1\right)$.

The impulse, $I$, is the same as the change in momentum, $D e < p$, and if the mass is the same before and after the collision, the change in momentum can be described as $m \Delta v$

$I = \Delta p = m \Delta v$

The change in the velocity is final minus initial, so $0 - 250 m {s}^{-} 1 = - 250 m {s}^{-} 1$

The negative number just means that the momentum has decreased, not increased, but we can ignore it in calculating the mass of the bullet. If we don't ignore it we end up with a negative mass, which doesn't make sense in this case.

$I = m \Delta v$

Rearranging to make $m$ the subject:

$m = \frac{I}{\Delta v} = \frac{5.0}{250} = 0.02 k g$