Answer:
1 kgm/s
250 J
10 N
1666.67\ N
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
[tex]p=m(v-u)\\\Rightarrow p=0.002(500-0)\\\Rightarrow p=1\ kgm/s[/tex]
Magnitude of the momentum of each pellet is 1 kgm/s
Kinetic energy
[tex]K=\frac{1}{2}m(v^2-u^2)\\\Rightarrow K=\frac{1}{2}\times 0.002(500^2-0^2)\\\Rightarrow K=250\ J[/tex]
Kinetic energy of each pellet is 250 Joules
Impulse
[tex]J=Ft[/tex]
[tex]J=m(v-u)\\\Rightarrow Ft=nm(v-u)\\\Rightarrow F=\frac{nm(v-u)}{t}\\\Rightarrow F=\frac{10\times 0.002\times (0-500)}{1}\\\Rightarrow F=-10\ N[/tex]
The magnitude of average force on the wall from the stream of pellets is 10 N
When t = 0.006 s
[tex]F=\frac{m(v-u)}{t}\\\Rightarrow F=\frac{0.002\times (0-500)}{0.006}\\\Rightarrow F=-1666.67\ N[/tex]
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
