Question

A small wooden block with mass 0.800 kg is suspended from the lower end of a light cord that is 1.44 m long. The block is initially at rest. A bullet with mass 0.0134 kg is fired at the block with a horizontal velocity v0. The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of 0.800 m , the tension in the cord is 4.76 N .

What was the initial speed v0 of the bullet?

Answer 1

298.04 m/s

Step-by-step explanation:

Let m = mass of bullet = 0.0134 kg and M = mass of block = 0.800 kg.

Since the bullet becomes embedded in the block and rises a vertical height,h = 0.800 m

The kinetic energy change of mass + block = potential energy of mass + block at height, h

ΔK = -ΔU

So, 1/2(m + M)(v² - V²) = -[(m + M)gh - 0] where v is the velocity of the bullet + block at height, h. Since the tension, T is the centripetal force at height, h, it follows that

T = (m + M)v²/r r = length of cord = 1.44 m

v = √(Tr/(m + M)) = √4.76 N × 1.44 m/(0.800 + 0.0134)kg = √(6.8544/0.8134) = √8.427 = 2.9 m/s

So. 1/2(v² - V²) = -gh

v² - V² = -2gh

V = √(v² + 2gh) = √((2.9 m/s)² + 2 × 9.8 m/s² × 0.8 m) = √(8.41 + 15.68) = √24.09 = 4.91 m/s

This is the velocity of the bullet plus block at collision.

From the law of conservation of momentum,

momentum of bullet = momentum of bullet plus block

mv₀ = (m + M)V where v₀ = initial speed of bullet

v₀ = (m +M)V/m = (0.0134 kg + 0.800 kg)4.91 m/s ÷ 0.0134 kg = 3.994 ÷ 0.0134 kg = 298.04 m/s

Answer 2

v₀ = 240 m / s

Step-by-step explanation:

This problem must be solved in two parts, first we must use the conservation of the moment, then the conservation of energy.

Let's start by applying moment conservation, to the system formed by the block and bullet, in this case the forces during the crash are internal and the moment is conserved

Instant starts. Before the crash

p₀ = m v₀

Final moment. Right after the crash

`p_(f)` = (m + M) v

The moment is preserved

Po =p_{f}

M v₀ = (m + M) v

v = m / (m + M) v₀ (1)

This is the speed with which the bullet block system comes out, now we can use energy conservation

Starting point. Right after the crash

Em₀ = K = ½ (m + M) v²

Final point. Highest point of the path

`Em_(f)` = U = (m + M) g y

Em₀ = Em_{f}

½ (m + M) v² = (m + M) g y

v = √2 g y (2)

We substitute 1 in 2

m / (m + M) v₀ = √ 2gy

v₀ = (m + M) / m √ 2gy

Let's calculate

v₀ = (0.0134 +0.800) /0.0134 √ (2 9.8 0.8)

v₀ = 240 m / s