Answer
The magnitude of the average force on the wall from each pellet is $~~1670~N$
Work Step by Step
In part (a), we found that the magnitude of the momentum of each pellet is $~~1.0~kg~m/s$
Therefore, the magnitude of the impulse on each pellet from the wall is $~~J = 1.0~kg~m/s$
We can find the magnitude of the average force on each pellet from the wall:
$F~t = J$
$F = \frac{J}{t}$
$F = \frac{1.0~kg~m/s}{0.60\times 10^{-3}~s}$
$F = 1670~N$
By Newton's Third Law, the average force on the wall from each pellet is equal in magnitude, but opposite in direction, to the average force on each pellet from the wall.
Therefore, the magnitude of the average force on the wall from each pellet is $~~1670~N$.
