A bullet (mass= 0.150 kg) moving horizontally at 470 m/s penetrates a 4.50 kg wo

Question

A bullet (mass= 0.150 kg) moving horizontally at 470 m/s penetrates a 4.50 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: KE_after = 1/2 m v^2 bullet + 1/2 M v^2 block you get less than the total KE before the impact: KE_before = 1/2 m v^2 bullet + 1/2 M v^2 block. 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,

(0.150 x 470) + (4.50 x 0) = (4.50 v) + (0.150 x 376)

v = 3.13 m/s

momentum of block = 4.50 x 3.13 = 14.1 kg m/s

(b) (0.150 x 470) + (4.50 x 0) = (4.50 v) + (- 0.150 x 376)

v = 28.2 m/s

momentum of block = 4.50 x 28.2 = 126.9 kg m/s

(c)

total mass = 4.65 kg

(0.150 x 470) = (0.150 + 4.50) v

v = 15.16 m/s ............Ans

momentum = 4.50 x 15.16 = 68.22kg 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 = 0.150 x 470^2 / 2 = 16567.5 K

KEf = 4.65 x 15.16^2 / 2 = 534.34 J

deltaKE = KEi - KEf = 16033.2 J

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