A 4.00 g bullet, traveling horizontally with a speed of 400 m/s, is fired into a

Question

A 4.00 g bullet, traveling horizontally with a speed of 400 m/s, is fired into a wooden block with mass 0.800 kg, initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 190m/s. (a) What is the speed of the block immediately after the impact? (b) If the coefficient of kinetic friction between block and surface is 0.46, how far does the block slide along the surface from its initial position before coming to rest? (Solve part (b) by using work and kinetic energy)

Explanation / Answer

here,
intital velocity of bullet, vi = 400 m/s
final velocity of bullet, vf = 190 m/s
mass of bullet, mb = 4 g = 0.004 kg
mass of block, m = 0.8 kg

From Conservation of momentum,
before collision = after collision

mb*vi = mb*vf + m*v

Solving for velocity of block after collision, v
v = (mb*vi - mb*vf)/m
v = (0.004*400 - 0.004*190)/0.8
v = 1.05 m/s towards bullet motion

Since work Done, w = force * Distance
w = frictional force * distance
w = uk*m*g*d

From Work energy theoram, w = Kinetic energy of block
uk*m*g*d = 0.5* m* v^2

Solving for distance travelled by block, d = v^2/(uk*g)
d = (1.05)^2/(0.46*9.81)
d = 0.244 m

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