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
