Respuesta :
Answer:1.084
Explanation:
Given
mass of Pendulum M=10 kg
mass of bullet m=5.5 gm
velocity of bullet u
After collision let say velocity is v
conserving momentum we get
[tex]mu=(M+m)v[/tex]
[tex]v=\frac{m}{M+m}\times u[/tex]
Conserving Energy for Pendulum
Kinetic Energy=Potential Energy
[tex]\frac{(M+m)v^2}{2}=(M+m)gh[/tex]
here [tex]h=L(1-\cos \theta )[/tex] from diagram
therefore
[tex]v=\sqrt{2gL(1-\cos \theta )}[/tex]
initial velocity in terms of v
[tex]u=\frac{M+m}{m}\times \sqrt{2gL(1-\cos \theta )}[/tex]
For first case [tex]\theta =6.8^{\circ} [/tex]
[tex]u_1=\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}[/tex]
for second case [tex]\theta =11.4^{\circ}[/tex]
[tex]u_2=\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}[/tex]
Therefore [tex]\frac{u_1}{u_2}=\frac{\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}}{\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}}[/tex]
[tex]\frac{u_1}{u_2}=\frac{1819.181\times 0.0838}{1001\times 0.1404}[/tex]
[tex]\frac{u_1}{u_2}=1.084[/tex]
i.e.[tex]\frac{v_1}{v_2}=1.084[/tex]
The conservation of momentum and energy allows finding the result for the relationship of the velocities of the balls fired by the two catapults is:
[tex]\frac{v_1}{v_2} = 1.0 85[/tex]
Given parameters
- The mass of the pendulum M = 10 kg.
- The mass of each ball m₁ = 5.50 g = 5.50 10⁻³ kg and m₂ = 10.0 g.
- pendulum angle te1 = 6.80º and tea2 = 11.4º
To find
- The ratio of ball speeds v1 / v2
The momentum is defined by the product of the massby the speed of the particle, without the system is isolated the momentum is conserved.
P = mv
Where bold letters indicate vectors, p is momentum, m is mass, and v is velocity.
Let's solve the problem for each ball separately, let's start with case 1 the ball of m₁ = 5.50 10⁻³ kg.
The system is shaped by the ball and the pendulum, therefore it is an isolated system and the momentum is conserved.
Initial instant. Before crash
p₀ = m₁ v₁
Final moment. Right after the crash
[tex]p_f[/tex] = (m₁ + M) v
The momentum is preserved
p₀ = [tex]p_f[/tex]
m₁ v₁ = (m₁ + M) v
v₁ = [tex]\frac{m_1 + M}{m_1} \ v[/tex] (1)
Now we can use the conservation of energy for the pendulum with the embedded ball m1, see attached.
Starting point. Just when they're coming out at the bottom
Em₀ = K = ½ (m₁ + M) v²
Final point. The highest part of the movement
[tex]Em_f[/tex]= U = (m₁ + M) gh
Energy is conserved.
Em₀ = [tex]Em_f[/tex]
½ (m₁ + M) v² = (m₁ + M) gh
v = [tex]\sqrt{2gh}[/tex]
Let's use trigonometry to find the height of the pendulum, see attached
h = L - L cos θ₁
Let's substitute
v = [tex]\sqrt{2gL ( 1 - cos \theta_1)}[/tex]
Let's substitute in equation 1
[tex]v_1 = \frac{m_1+M}{m_1} \ \sqrt{2g L ( 1- cos \theta_1) }[/tex]
We solve the case 2 ball of mass m₂ = 10 10⁻³ kg.
The system has the same approach: the conservation of the moment and then the conservation of energy, therefore the expression for the speed of this ball is:
[tex]v_2 = \frac{m_2+M}{m_2} \ \sqrt{2gL(1-cos \theta_2) }[/tex]
They ask us for the relationship between the two speeds.
Let's calculate each velocity and then make the relationship.
Suppose a pendulum length value of L = 1 m, to make an explicit calculation.
[tex]v_1 = \frac{0.0055 + 10}{0.0055} \ \sqrt{2 \ 9.8 \ 1 (1 - cos 6.8)}[/tex]
v₁ = 675.49 m / s
[tex]v_2 = \frac{0.01 + 10}{0.010} \ \sqrt{2 \ 9.8 \ 1(1 - cos 11.4)}[/tex]
v₂ = 622.46 m/s
The ratio of the speeds is:
[tex]\frac{v_1}{v_2} = \frac{675.49}{622.46}[/tex]
[tex]\frac{v_1}{v_2}[/tex] = 1,085
In conclusion, using the conservation of momentum and energy, we can find the relationship of the velocities of the balls fired by the two catapults is:
[tex]\frac{v_1}{v_2}[/tex] = 1,085
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