A 20g bullet moving at 200 meter per second hits a bag of sand and comes to rest in 0.011secs.What is the momentum of the bullet just before hitting the bag?Find the average force that stopped the bullet.
momentum = mass * velocity ... kg•m/s
acceleration = v / t = 200 m/s / 11 ms ... m/s^2
force = mass * acceleration = .02 kg * (200 m/s / 11 ms) ... N
To find the momentum of the bullet just before hitting the bag, we can use the formula:
Momentum = mass × velocity
Given that the mass of the bullet is 20 grams (or 0.02 kg) and its velocity is 200 meters per second, we can substitute these values into the formula to calculate the momentum:
Momentum = 0.02 kg × 200 m/s = 4 kg*m/s
Hence, the momentum of the bullet just before hitting the bag is 4 kg*m/s.
To find the average force that stopped the bullet, we can use the formula:
Average Force = change in momentum / change in time
The change in momentum is given by the initial momentum (4 kg*m/s) minus the final momentum (which is zero since the bullet comes to rest in the bag). The change in time is given as 0.011 seconds.
Substituting these values into the formula, we have:
Average Force = (0 kg*m/s - 4 kg*m/s) / 0.011 s
= - 4 kg*m/s / 0.011 s
Hence, the average force that stopped the bullet is approximately -363.64 N (since the force is opposing the motion, the negative sign indicates its direction).
Note: The negative force value indicates that the force was applied in the opposite direction of the bullet's initial motion, causing it to come to rest.