Question

A bullet of inertia m traveling at speed v is fired into a wooden block that has inertia 4m and rests on a level surface. The bullet passes through the block and emerges with speed v/3, taking a negligible amount of the wood with it. The block moves to the right but comes to rest after traveling a distance d.
What is the magnitude of the frictional force between the block and the surface while the block is moving?


What is the ratio of the energy dissipated as the bullet passes through the block to the energy dissipated by friction between the surface and the bottom face of the block?


I solved the first one and got (4mv^2)/(9d) but it said "your answer either contains an incorrect numerical multiplier or is missing one" so I'm not really sure what I did wrong.

Answer 1

Answer:

F = mv²/(18d)

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Explanation:

Momentum before collision = momentum after collision

mv = m(v/3) + (4m) V

v = v/3 + 4V

4V = 2v/3

V = v/6

The acceleration of the block is:

v² = v₀² + 2aΔx

(0 m/s)² = (v/6)² + 2ad

2ad = -v²/36

a = -v²/(72d)

The friction force is therefore:

∑F = ma

-F = (4m) (-v²/(72d))

F = mv²/(18d)

The energy dissipated during the collision is:

ΔE = 1/2 mv² − (1/2 m(v/3)² + 1/2 (4m)(v/6)²)

ΔE = 1/2 mv² − (1/18 mv² + 1/18 mv²)

ΔE = 1/2 mv² − 1/9 mv²

ΔE = 7/18 mv²

The energy dissipated by the friction force:

W = Fd

W = 1/18 mv²

The ratio is:

ΔE / W

(7/18 mv²) / (1/18 mv²)

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