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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Orla Orla 2 months 2021-07-31T14:51:30+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-07-31T14:52:54+00:00

    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

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