A bullet (mass= 0.180 kg) moving horizontally at 470 m/s penetrates a 4.00 kg wo

Question

A bullet (mass= 0.180 kg) moving horizontally at 470 m/s penetrates a 4.00 kg wood block resting on a frictionless table.

a) After emerging from the block, the bullet has been slowed down to 376 m/s.

What is the final speed of the block? _____ m/s.
Momentum Total after Impact: ______ kg m/s.

b) Repeat the above question, only now assume that the bullet bounces back at 376 m/s.

What is the final speed of the block? _____m/s.
Momentum Total after Impact: ______kg m/s.

c) Repeat the above question, only now figure that the bullet sticks to the block.

What is the mass of the (block+bullet)? _____ kg.
What is the final speed of the block? ______ m/s.
Momentum Total after Impact: _______ kg m/s.

In all of these cases, if you compute the total KE after the impact:

KEafter = 1/2 m v2bullet + 1/2 M v2block

you get less than the total KE before the impact:

KEbefore = 1/2 m v2bullet + 1/2 M v2block .

Find the case where this KE changes the most.
At the most, the total KE changes by: _____ J. (Answer as a positive value).

Explanation / Answer

(a) Applying momentum conservation for the collision,

(0.180 x 470) + (4 x 0) = (4v) + (0.180 x 376)

v = 4.23 m/s ......Ans

momentum of block = 4 x 4.23 = 16.92 kg m/s .....Ans

(b) (0.180 x 470) + (4 x 0) = (4 v) + (- 0.180 x 376)

v = 38.07 m/s ...Ans

momentum of block = 4 x 38.07 =152.28 kg m/s ..Ans

(c) total mass = 4.18 kg ......Ans

(0.180 x 470) = (4.18) v

v = 20.24 m/s ............Ans

momentum = 4 x 20.24 = 81 kg m/s .......Ans

(d) KE changes most in case (C)

because this is perfectly inelastic collision.

and energy lost is maximum in this case.

KEi = 0.180 x 470^2 / 2 = 19881 K

KEf = 4.18 x 20.24^2 / 2 = 856.18 J

deltaKE = KEi - KEf = 19024.8 J

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