Question

A small wooden block with mass 0.775 kg is suspended from the lower end of a light cord that is 1.50 m long. The block is initially at rest. A bullet with mass 0.0120 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 0.725 m, the tension in the cord is 4.88 N.
What was the initial speed v0 of the bullet?

Answer 1

Answer:

34.83 m/s

Explanation:

From the law of conservation of momentum,

initial momentum of bullet = final momentum of block + bullet

mv₀ = (m + M)V

V = mv₀/(m + M)

where m = mass of bullet = 0.0120 kg, v₀ = initial momentum of bullet, M = mass of block = 0.775 kg, V = final velocity of block + bullet.

Now, since the block + bullet rise a height of 0.725 m, from the law of conservation of energy,

potential energy change of block + bullet = kinetic energy change of block + bullet.

So (m + M)gh - 0 = -1/2(m + M)(V₁² - V²) where h = vertical height moved = 0.725 m and V₁ = velocity at 0.725 m and it has zero potential energy initially.

gh = -1/2(V₁² - V²)   (2)

Now, we obtain V₁ from

F = (m + M)V₁²/R since a centripetal force acts on the block + bullet at height 0.725 m. F = tension in chord = 4.88 N and R = length of cord = 1.50 m.

V₁ = √[FR/(m + M)]

Substituting V and V₁ into (2) above, we get

gh = -1/2(FR/(m + M) - [mv₀/(m + M)]²)

-2(m + M)²gh = FR(m + M) - (mv₀)²

v₀ = √([FR(m + M) + 2(m + M)²gh]/m)

substituting the values of the variables into v₀ we have

v₀ = √([4.88 N × 1.50 m × (0.0120 kg + 0.775 kg)  + 2(0.0120 kg + 0.775 kg)² × 9.8 m/s² × 0.725 m]/0.0120 kg)

= √([7.32 × 0.787 + 2(0.787)² × 9.8 m/s² × 0.725 m]/0.0120 kg)

= √(5.76 + 8.80)/0.012 kg

= √14.56/0.012

= √1213.40

= 34.83 m/s

So the initial speed v₀ = 34.83 m/s