Question

A pellet gun fires ten 2.0 g pellets per second with a speed of 500 m/s. The pellets are stopped by a rigid wall. What are

(a) the magnitude of the momentum of each pellet,
(b) the kinetic energy of each pellet, and
(c) the magnitude of the average force on the wall from the stream of pellets?
(d) If each pellet is in contact with the wall for 0.60 ms, what is the magnitude of the average force on the wall from each pellet during contact?
(e) Why is this average force so different from the average force calculated in (c)?

Answer 1

1 kgm/s

250 J

10 N

1666.67\ N

Step-by-step explanation:

t = Time taken

u = Initial velocity = 0 (assumed the gun was at rest)

v = Final velocity = 500 m/s

n = Number of pellets/ second = 10

m = Mass of pellet = 0.002 kg

F = Force

Momentum

`p=m(v-u)\\\Rightarrow p=0.002(500-0)\\\Rightarrow p=1\ kgm/s`

Magnitude of the momentum of each pellet is 1 kgm/s

Kinetic energy

`K=(1)/(2)m(v^2-u^2)\\\Rightarrow K=(1)/(2)* 0.002(500^2-0^2)\\\Rightarrow K=250\ J`

Kinetic energy of each pellet is 250 Joules

Impulse

`J=Ft`

`J=m(v-u)\\\Rightarrow Ft=nm(v-u)\\\Rightarrow F=(nm(v-u))/(t)\\\Rightarrow F=(10* 0.002* (0-500))/(1)\\\Rightarrow F=-10\ N`

The magnitude of average force on the wall from the stream of pellets is 10 N

When t = 0.006 s

`F=(m(v-u))/(t)\\\Rightarrow F=(0.002* (0-500))/(0.006)\\\Rightarrow F=-1666.67\ N`

The magnitude of the average force on the wall from each pellet during contact is 1666.67 N

Part c was the cumulative force of all the pellet while part d was the force of each pellet. Hence, the difference