15. A pellet gun fires ten 2.9 g pellets per second with a speed of 580 m/s . Th

Question

15. A pellet gun fires ten 2.9 g pellets per second with a speed of 580 m/s . The pellets are stopped by a rigid wall.

(a) What is the momentum of each pellet?  

(b) What is the kinetic energy of each pellet?

(c) What is the average force exerted by the stream of pellets on the wall?

(d) If each pellet is in contact with the wall for 0.60ms, what is the average force exerted on the wall by each pellet during contact? Why is this average force so different from the average force calculated in (c)?

Explanation / Answer

a) momentum = mv = (2,9 x 10^-3) x 580 = 1.682 Kg.m/s

b) K.E. = mv^2 /2 = (2.9 x 10^-3) x 580^2 /2 = 487.78 J

c) To calulate Average force time of contact needed.

d) Impulse = change in momentum

Impulse = F x t

1.682   = F x 0.60 x 10^-3

F = 2803.33 N

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