A 20g bullet moving at 200m/s hits a bag of Sand and comes to rest in 0.011s what is the momentum of the bullet just before hitting the bag? Find the average Force that stopped the bullet
Well, I've never been a big fan of bullet stopping contests, but I'll give it a shot! First, let's calculate the momentum of the bullet just before it hit the bag.
Momentum (p) is calculated by multiplying the mass (m) of an object by its velocity (v). In this case, the bullet has a mass of 20 grams (or 0.02 kg) and a velocity of 200 m/s.
p = m * v
p = 0.02 kg * 200 m/s
p = 4 kg m/s
So, the momentum of the bullet just before hitting the bag is 4 kg m/s. Now, let's move on to finding the average force that stopped the bullet.
To find the average force (F), we can use the formula F = Δp / Δt, where Δp is the change in momentum and Δt is the time it took for the bullet to come to a stop. In this case, the change in momentum is the final momentum of the bullet (which is 0 kg m/s, since it comes to rest) minus the initial momentum (which is 4 kg m/s), and the time is 0.011 seconds.
F = Δp / Δt
F = (0 kg m/s - 4 kg m/s) / 0.011 s
F = -4 kg m/s / 0.011 s
F ≈ -363.64 N
So, the average force that stopped the bullet is approximately -363.64 Newtons. Keep in mind that the negative sign indicates that the force is acting opposite to the direction of the bullet's initial motion.
Now, if you'll excuse me, I'll go find a more cheerful topic to clown around with!