Explain why a nerf gun is safe using momentum (assuming it's an elastic collision).

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The question gives a great clue to the answer. Here we start with a system (Nerf gun and dart) that is at rest with velocity 0. Given the equation for basic momentum (`p=mv` ), the total momentum of the system must also be zero. This is probably the most important concept to understand this question, along with the idea of conservation of momentum, which tells us that unless something changes (it doesn't in the setting of the problem) the total momentum of the system will always be 0.

Momentum is additive as in the following equation:

`p_(system) = p_(gun) + p_(dart)=0`

Therefore, we can relate the masses and velocities of the gun and dart once the trigger is pulled:

`m_(gun)*v_(gun) = -m_(dart) * v_(dart)`

The negative is only there because the velocity vectors are going to be in opposite directions (and we can't have negative mass in classic physics problems!).

Now, let's rearrange the equation:

`- (v_(gun)/v_(dart)) = m_(dart)/m_(gun)`

You can see that the ratio of the velocities is the inverse of the ratio of the masses of the different objects. It is difficult to find an official or academic source on the masses of nerf guns and darts, but the sources linked suggest that the darts are about 1 gram each, and guns range from 180 grams - 6 pounds. Let's take the "worst case" using the lightest—180 gram—gun (we'll also get rid of the negative sign and discuss in terms of speeds, which are the absolute values of velocity):` `

`s_(gun)/s_(dart) = 1/180`

We can see that the speed of the gun is very low compared to the speed of the dart. Similarly, the kinetic energy will be the same fraction of the kinetic energy of the dart. Therefore, not only is the gun much slower, but it carries significantly less energy to be absorbed by the shooter!

We might be able to do the same calculation by using conservation of energy, too, but we don't have enough information for that analysis. We would need to be able to calculate the potential energy of the system (the tension in the spring of the gun primarily) to estimate the final energies of the gun and dart.

Last Updated by eNotes Editorial on December 26, 2019
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