Question

A 4.50-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

a) μ = 0.1957 , b) ΔK = 158.8 J , c) K = 0.683 J

Step-by-step explanation:

We must solve this problem in parts, one for the collision and the other with the conservation of energy

Let's find the speed of the wood block after the crash

Initial moment. Before the crash

p₀ = m v₁₀ + M v₂₀

Final moment. Right after the crash

pf = m
`v_(1f)` + M v_{2f}

The system is made up of the block and the bullet, so the moment is preserved

p₀ = pf

m v₁₀ = m v_{1f} + M v_{2f}

v_{2f} = m (v₁₀ - v_{1f}) / M

v_{2f} = 4.5 10-3 (400 - 190) /0.65

v_{2f} = 1.45 m / s

Now we can use the energy work theorem for the wood block

Starting point

Em₀ = K = ½ m v2f2

Final point

Emf = 0

W = ΔEm

- fr x = 0 - ½ m v₂₂2f2

The friction force is

fr = μN

With Newton's second law

N- W = 0

N = Mg

We substitute

-μ Mg x = - ½ M v2f2

μ = ½ v2f2 / gx

Let's calculate

μ = ½ 1.41 2 / 9.8 0.72

μ = 0.1957

b) let's look for the initial and final kinetic energy

K₀ = 1/2 m v₁²

K₀ = ½ 4.50 10⁻³ 400²

K₀ = 2.40 10² J

Kf = ½ 4.50 10⁻³ 190²

Kf = 8.12 10¹ J

Energy reduction is

K₀ - Kf = 2.40 10²- 8.12 10¹

ΔK = 158.8 J

c) kinetic energy

K = ½ M v²

K = ½ 0.650 1.45²

K = 0.683 J