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The figure below shows a bullet of mass 190 g traveling horizontally towards the

ID: 778847 • Letter: T

Question

The figure below shows a bullet of mass 190 g traveling horizontally towards the east with speed 415 m/s, which strikes a block of mass 8.0 kg that is initially at rest on a frictionless table.

What are the magnitude (in m/s) and direction of the velocity of the block/bullet combination immediately after the impact?

magnitude (in m/s)

direction: (north, south, east, west, up, down, zero)?

(b)

What are the magnitude (in N · s) and direction of the impulse by the block on the bullet?

magnitude (in N · s)

direction: (north, south, east, west, up, down, zero)?

(c)

What are the magnitude (in N · s) and direction of the impulse from the bullet on the block?

magnitude (in N · s)

direction: (north, south, east, west, up, down, zero)?

(d)

If it took 3 ms for the bullet to change the speed from 415 m/s to the final speed after impact, what is the average force (in N) between the block and the bullet during this time? (Enter the magnitude in N.)

Explanation / Answer

m1(bullet) = 190 g = 0.19 kg          m2(block) = 8 kg

before collision

speed of bullet v1i = 415 m/s

speed of blovk v2i = 0

momentum before collision Pi = m1*v1i + m2*v2i

after collision


the block and bullet combines

speed of bullet + block = vf

momentum after collision Pf = (m1+m2)*vf


from momentum conservation


total momentum is conserved

momentum before collision = momentum after collision


Pi = Pf

(0.19*415) + 0 = (0.19 + 8)*vf

vf = 9.627 m/s

direction east


====================================


(b)

the magnitude of the impulse by the block on the bullet = m1*vf - m1*v1i

the magnitude of the impulse by the block on the bullet = 0.19*9.627 - 0.19*415 = -77.02 N s

direction = West

(c)

the magnitude of the impulse from the bullet on the block = m2*vf - m2*v2i = 8*9.627 = 77.02 Ns


direction East

-----------------------

(d)


average force = impulse*time = 77.02*3*10^-3 = 0.231 N

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