A bullet of mass m is fired into a block of mass M that is at rest. The block, w

Question

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 muk. Part A Find an expression for the bullet?s speed vbullet. Express your answer in terms of the variables m, M, muk, d, and appropriate constants. Part B What is the speed of a 11 g bullet that, when fired into a 12 kg stationary wood block, causes the block to slide 6.0 cm across a wood table? Assume that muk = 0.20. Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Here ,

let the speed of block just after bullet embedding is vB

accelertion due to friction = - uk * g

Using third equation of motion for the motion of block

0 - vB^2 = -2 * uk * g * d

vB = sqrt(2*uk*g*d)

Now , for the block -bullet collision ,

using conservation of momentum

m * vbullet = (M + m) * sqrt(2*uk*g*d)

vbullet = (M + m) * sqrt(2*uk*g*d)/m

the velocity of bullet is (M + m) * sqrt(2*uk*g*d)/m

part B)

putting values is equation

vbullet = (0.011 + 12) * sqrt(0.2 * 9.8 * 2 *0.06)/.011

vbullet = 529.5 m/s

the speed of bulelt is 529.5 m/s

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

Here ,

as the coefficient of friction is same for both

acceleration will also be same

Now , for the same acceleration

v1^2/2d1 = v2^2/2d2

for the blocks , as one is 7 times massive than the other

7 * v1 = v2

d1 = 6.7 m

v1^2/6.7 = 7^2 * v1/d2

d2 = 328.3 m

the lighter block will slide 328.3 m

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