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In a ballistics test, a 25 g bullet traveling horizontally at 1200m/s goes throu

ID: 1681649 • Letter: I

Question

In a ballistics test, a 25 g bullet traveling horizontally at 1200m/s goes through a 25 cm thick 350 kgstationary target and emerges with a speed of 700 m/s. The target is free to slide on a smoothhorizontal surface. (a) How long is the bullet in the target?
1 s I got this right
What average force does it exert on the target?
2 N (magnitude only) I got thisright
(b) What is the target's speed just after the bullet emerges?
    m/s (a) How long is the bullet in the target?
1 s I got this right
What average force does it exert on the target?
2 N (magnitude only) I got thisright
(b) What is the target's speed just after the bullet emerges?
    m/s

Explanation / Answer

For the third part , we can calculate it by using law ofconservation of momentum , Initial momentum when bullet was fired , Momentum = momentum of bullet + momentum of block                  =0.025*1200 + 350*0 (momentum = mass*velocity)                  =30 kgm/s Final momentum as soon at it emerges out of block = momentumof bullet + momentum of block                  =0.025*700 + 350*v (v is the final velocity of block)                  =17.5 + 350v kgm/s Equating , 30 = 17.5 + 350 v => v = 3.57 cm/s = 3.57*10-2 m/s