Respuesta :
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
The velocity of the bullet can be found by using the laws of conservation of
energy and momentum.
The velocity of the bullet is approximately 318.1 m/s
Reasons:
Given parameters are:
Mass of block = 0.775 kg
Length of cord = 1.50 m
Mass of the bullet = 0.0120 kg
Velocity of bullet = v₀
Height to which the block rises = 0.725 m
Tension in the chord = 4.88 N
Required:
Initial speed of the bullet, v₀
Solution:
The conservation of momentum equation is 0.0120·v₀ = (0.775 + 0.0120)·v
Centripetal force = Tension in string = [tex]\dfrac{(m + M) \cdot v_c^2}{l} = F[/tex]
Where;
[tex]v_c[/tex] = The velocity at the height, h, applying centripetal force (tension) in the
cord
Therefore;
[tex]\dfrac{(0.0120 + 0.775) \times v_c^2}{1.50} = 4.88[/tex]
[tex]v_c = \sqrt{ \dfrac{4.88 \times 1.50}{(0.0120+0.775)} } \approx 3.05[/tex]
Change in kinetic energy = Gain in potential energy
[tex]\dfrac{1}{2} \cdot (m + M) \cdot (v^2 - v^2_c) = (m + M) \cdot g \cdot h[/tex]
[tex](m + M) \cdot g \cdot h + \dfrac{1}{2} \cdot (m + M) \cdot v^2_c = \dfrac{1}{2} \cdot (m + M) \cdot v^2[/tex]
[tex]g \cdot h + \dfrac{1}{2} \cdot v^2_c = \dfrac{1}{2} \cdot v^2[/tex]
[tex]\dfrac{4.88 \times 1.50}{(0.0120+0.775)} + 2 \times 9.81 \times 0.725 = v^2[/tex]
v ≈ √(23.53) ≈ 4.85
v ≈ 4.85 m/s
0.0120·v₀ = (0.775 + 0.0120)·v
Therefore;
[tex]v_0 = \dfrac{(0.775 + 0.0120) \cdot v}{0.0120}[/tex]
[tex]v_0 = \dfrac{(0.775 + 0.0120) \times 4.85}{0.0120} \approx 318.1[/tex]
The velocity of the bullet, v₀ ≈ 318.1 m/s
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