A pellet gun fires ten 2.14-g pellets per second with a speed of 483 m/s. The pellets are stopped by a rigid wall. (a) Find the momentum of each pellet. (b) Calculate the average force exerted by the stream of pellets on the wall. (c) If each pellet is in contact with the wall for 1.25 ms, what is the average force exerted on the wall by each pellet while in contact? Why is this so different from (b)?

(a) To obtain the momentum p of each pellet, substitute 2.14-g for mass m, 483 m/s for v in the equation p = mv,p = mv
= (2.14-g) (483 m/s)
= (2.14-g×10^-3 kg/1 g) (483 m/s)
= 1.03 kg.m/s
Therefore the momentum p of each pellet would be 1.03 kg.m/s.
(b) As a pellet gun fires ten 2.14-g pellets per second with a speed of 483 m/s, thus the impulse J imparted to the wall in one second is ten times of the momentum p of each pellet.
So, J = 10p
= 10(1.03 kg.m/s)
= 10.3 kg.m/s