This is the second time I am submitting this problem. Please do not give me the

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This is the second time I am submitting this problem. Please do not give me the same answer that I got previously, as shown in the answer box of the problem screenshot.

Problem 11.49 Part A Find an expression for the bullet's speed vbullet A bullet of mass m is fired into a block of mass M that is at rest. The block, with the bullet embedded slides distance d across a horizontal surface. The coefficient of kinetic friction is k Express your answer in terms of the variables m, Mg, d, and appropriate constants Ubullet Submit My Answers Give D Incorrect Iry Again; no points deducted Part B What is the speed of a 12 g bullet that, when fired into a 12 kg stationary wood block, causes the block to slide 4.8 cm across a wood table? Assume that = 0.20 Express your answer to two significant figures and include the appropriate units |Im bullet =113.72 Submit My Answers Give D Incorrect Iry Again Continue

Explanation / Answer

A)

consider the motion after collision of bullet and block

Vi = initial velocity

Vf = final velocity = 0 m/s

using conservation of energy

work done by kinetic frictional force = change in kinetic energy

- fk d = (0.5) m (Vf2 - Vi2)

- uk (M + m)g d =  (0.5) (M + m) ((0)2 - Vi2)

Vi = sqrt(2uk gd)

consider the collision between bullet and block

v = velocity of the bullet

using conservation of momentum

mv = (m + M) Vi

v = ((m + M)/m)sqrt(2uk gd)

b)

m = 12 g = 0.012 kg

M = 12 kg

uk = 0.20

d = 4.8 cm = 0.048 m

v = ((m + M)/m)sqrt(2uk gd)

v = ((0.012 + 12)/0.012)sqrt(2(0.20) (9.8) (0.048))

v = 434.21 m/s

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