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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
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
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Physics
4 years
2021-07-31T14:51:30+00:00
2021-07-31T14:51:30+00:00 1 Answers
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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