Question

A 4.00-g bullet, traveling horizontally with a velocity of magnitude 400 m/s, is fired into a wooden block with mass 0.650 kg , initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 190 m/s. The block slides a distance of 72.0 cm along the surface from its initial position.
a. What is the coefficient of kinetic friction between block and surface?
b. What is the decrease in kinetic energy of the bullet?
c. What is the kinetic energy of the block at the instant after the bullet passes through it?

Answer 1

Answer:

a) Coefficient of kinetic friction between block and surface = 0.12

b) Decrease in kinetic energy of the bullet = 247.8 J

c) Kinetic energy of the block at the instant after the bullet passes through it = 0.541 J

Explanation:

Given,

Mass of bullet = 4.00 g = 0.004 kg

Initial velocity of the bullet = 400 m/s

Mass of wooden block = 0.65 kg

Initial velocity of the wooden block = 0 m/s (since it was initially at rest)

Final velocity of the bullet = 190 m/s

Distance slid through by the block after the collision = d = 72.0 cm = 0.72 m

Let the velocity of the wooden block after collision be v

According to the law of conservation of momentum,

Momentum before collision = Momentum after collision

Momentum before collision = (Momentum of bullet before collision) + (Momentum of wooden block before collision)

Momentum of bullet before collision = (0.004×400) = 1.6 kgm/s

Momentum of wooden block before collision = (0.65)(0) = 0 kgm/s

Momentum after collision = (Momentum of bullet after collision) + (Momentum of wooden block after collision)

Momentum of bullet after collision = (0.004×190) = 0.76 kgm/s

Momentum of wooden block after collision = (0.65)(v) = (0.65v) kgm/s

Momentum balance gives

1.6 + 0 = 0.76 + 0.65v

0.65v = 1.6 - 0.76 = 0.84

v = (0.84/0.65)

v = 1.29 m/s

The velocity of the wooden block after collision = 1.29 m/s

To obtain the coefficient of kinetic friction between block and surface, we will apply the work-energy theorem.

The work-energy theorem states that the work done in moving the block from one point to another is equal to the change in kinetic energy of the block between these two points.

The points to consider are the point when the block starts moving (immediately after collision) and when it stops as a result of frictional force.

Mathematically,

W = ΔK.E

W = workdone by the frictional force in stopping the wooden block (since there is no other horizontal force acting on the block)

W = -F.d (minus sign because the frictional force opposes motion)

d = Distance slid through by the block after the collision = 0.72 m

F = Frictional force = μN

where N = normal reaction of the surface on the wooden block and it is equal to the weight of the block.

N = W = mg

F = μmg

W = - μmg × d = (-μ)(0.65)(9.8) × 0.72 = (-4.59μ) J

ΔK.E = (final kinetic energy of the block) - (initial kinetic energy of the block)

Final kinetic energy of the block = 0 J (since the block comes to a rest)

(Initial kinetic energy of the block) = (1/2)(0.65)(1.29²) = 0.541 J

ΔK.E = 0 - 0.541 = - 0.541 J

W = ΔK.E

-4.59μ = -0.541

μ = (0.541/4.59)

μ = 0.12

b) The decrease in kinetic energy of the bullet

(Decrease in kinetic energy of the bullet) = (Kinetic energy of the bullet before collision) - (Kinetic energy of the bullet after collision)

Kinetic energy of the bullet before collision = (1/2)(0.004)(400²) = 320 J

Kinetic energy of the bullet after collision = (1/2)(0.004)(190²) = 72.2 J

Decrease in kinetic energy of the bullet = 320 - 72.2 = 247.8 J

c) Kinetic energy of the block at the instant after the bullet passes through it = (1/2)(0.65)(1.29²) = 0.541 J

Hope this Helps!!!