Question

A 200-gr (7000 gr = 1 lb) bullet goes from rest to 3300 ft/s in 0.0011 s. Determine the magnitude of the impulse imparted to the bullet during the given time interval. In addition, determine the magnitude of the average force acting on the bullet.

Answer 1

Answer:

The force acting on the bullet $F = 84000 \frac{lb ft}{s^{2} }$

The value of impulse on the bullet in the given time interval P = 92.4 $\frac{lb ft}{sec}$

Explanation:

Mass of the bullet ( m ) = 200 gr = 0.028 lb

Initial velocity ( U ) = 0

Final Velocity ( V ) = 3300 $\frac{ft}{sec}$

Force acting on the bullet $F = \frac{m ( V - U )}{t}$

⇒ $F = \frac{ 0.028 ( 3300 - 0 )}{0.0011}$

⇒ $F = 84000 \frac{lb ft}{s^{2} }$

This is the force acting on the bullet.

Magnitude of the impulse imparted on the bullet  $P = F dt$  -------- (1)

Put the value of F & dt in above equation we get,

P = 84000 × 0.0011

P = 92.4 $\frac{lb ft}{sec}$

This is the value of impulse on the bullet in the given time interval.