Question

A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average force (in N) exerted on a 0.0200 kg bullet to accelerate it to a speed of 550 m/s in a time of 3.50 ms (milliseconds)? (Enter the magnitude.)

Answer 1

The average force is 3142.86 N

Step-by-step explanation:

We would apply the formula for calculating rate of change of momentum which is expressed as

F = m(v - u)/t

where

F is the force

m is the mass of the object

v - u is the change in velocity

t is the time

From the information given,

m = 0.02

v - u = 550

t = 3.5ms

We would convert from ms to s. Recall,

1 ms = 0.001 s

3.5ms = 3.5 x 0.001 = 0.0035

Thus,

F = 0.02(550)/0.0035

F = 3142.86 N