Question

In a ballistics test, a 23 g bullet traveling horizontally at 1100 m/s goes through a 33-cm-thick 400 kg stationary target and emerges with a speed of 950 m/s . The target is free to slide on a smooth horizontal surface.

(a) how long is the bullet in the target?

______S

what average force does it exert on the target?

____N (magnitude only)

(b) What is the targets speed just after the bullet emerges?

_____m/s

Answer 1

Answer:

Explanation:

loss of energy while passing through target by bullet

= 1/2 mu² - 1/2 mv² , m is mass of bullet , u is initial velocity and v is final velocity .

= 1/2 x m ( u² - v² )

= .5 x .023 x ( 1100² - 950² )

= 3536.25 J.

This loss is due to negative work done by  friction force

If friction force be F

Work done by friction force = F x .33

F x .33 = loss of kinetic energy

F x .33 = 3536.25

F = 10716 N

impulse of F

F X t , time period during which this force remains active

10716 x t = change in momentum of bullet

= .023 ( 1100 - 950 )

= 3.45

t = 3.45 / 10716

= 3.22 x 10⁻⁴ s.

Average force on the target = friction force created = 10716 N

Impulse by force on target = 10716 x 3.22 x 10⁻⁴

impulse on target = change in momentum of target

= mass of target x its velocity after impact

=  400 v

v = 10716 x 3.22 x 10⁻⁴ / 400

= 86.26 x 10⁻⁴ m /s