9.43 A bullet of mass m is fired into a block of mass M that is at rest. The blo
ID: 1393759 • Letter: 9
Question
9.43
A bullet of mass m is fired into a block of mass Mthat is at rest. The block, with the bullet embedded, slides distance d across a horizontal surface. The coefficient of kinetic friction is ?k.
A. Find an expression for the bullet's speed vbullet.
Express your answer in terms of the variables m, M, ?k, d, and appropriate constants.
B. What is the speed of a 11g bullet that, when fired into a 9.0kg stationary wood block, causes the block to slide 4.8cm across a wood table? Assume that ?k=0.20.
Express your answer to two significant figures and include the appropriate units.
Explanation / Answer
Mass of the bullet m = m
Mass of the block M = M
Initial velocity of the bullet u = v
Initial velocity of the block U = 0
Sliding distance S = d
The coefficient of kinetic friction = u k
From law of conservation of linear momentum , mu + MU = ( m + M ) V
Substitute values you get mv + 0 = ( m+ M) V
Velocity of the bullet block system after collision V = mv / ( m+ M ) ----------( 1)
From kinematics equation , V ' 2 - V 2 = 2aS
Where V ' = 0 Since after sliding d distance it comes to rest .
a = accletation = -u k g
Substitute values you get -V 2 = 2(-u k g)d
V = (2u k g d) 1/2 -------( 2)
From equation ( 1) and ( 2) ,
mv / ( m+ M ) =(2u k g d) 1/2
Bullet speed v = [ (m+ M) / m](2u k g d) 1/2 ------( 3)
(B).Given m = 11 g = 11 x 10 -3 kg
M = 9 kg
d = 4.8 cm = 4.8 x 10 -2 m
u k = 0.2
Substitute these values in equation ( 3) you get ,
v = [ 819.1818 ] (0.18816) 1/2
= 819.1818 x 0.43377
= 355.33 m / s
= 360 m / s for two significant figures
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