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2 years ago

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In a ballistics test, a 24 g bullet traveling horizontally at 1300 m/s goes through a 25-cm-thick 360 kg stationary target and e

merges with a speed of 930 m/s. The target is free to slide on a smooth horizontal surface. How long is the bullet in the target? What average force does it exert on the target? What is the target's speed just after the bullet emerges?

1 answer:

MaRussiya [10]2 years ago

8 0

Answer:

A)0.00022s b)40363.6N c) 0.025m/s

Explanation:

Mass = 24g = 0.024kg, distance though the target = thickness of the target = 25cm = 0.25m

Initial speed of the bullet = 1300m/s, final speed = 930m/s

Using equation of motion

Distance = 1/2(vf+vi)*t (time in seconds)

t = 0.25*2/(1300+930) = 0.00022s

B) force exerted on the body

F = ma = m* (vf-vi)/t = 0.024*(930-1300)/0.00022

F = -40363N, it is negative because the body decelerated during this motion

C) using law of conservation of momentum,

M1*U1+ M2*U2(M2and U1 are the mass and initial speed of the body) = M1V1+ M2V2

The target was at rest so initial speed U2 = 0

0.024*1300 + 360*0 = 0.024*930 + 360*V2

31.2 = 22.32+360*V2

31.2-22.33 = 360*V2

V2 = 8.88/360 = 0.025m/s

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To solve this exercise, it is necessary to apply the concepts of conservation of the moment especially in objects that experience an inelastic colposition.

They are expressed as,

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Once the final velocity is found it is possible to calculate the change in kinetic energy, so

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Explanation:

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