Respuesta :
Answer:
h = 13.3 m
Explanation:
Given:-
- The mass of ball, mb = 4.80 kg
- The mass of bar, ml = 7.0 kg
- The height from which ball dropped, H = 15.0 m
- The length of bar, L = 6.0 m
- The mass at other end of bar, mo = 5.10 kg
Find:-
The dropped ball sticks to the bar after the collision.How high will the other ball go after the collision?
Solution:-
- Consider the three masses ( 2 balls and bar ) as a system. There are no extra unbalanced forces acting on this system. We can isolate the system and apply the principle of conservation of angular momentum. The axis at the center of the bar:
- The angular momentum for ball dropped before collision ( M1 ):
M1 = mb*vb*(L/2)
Where, vb is the speed of the ball on impact:
- The speed of the ball at the point of collision can be determined by using the principle of conservation of energy:
ΔP.E = ΔK.E
mb*g*H = 0.5*mb*vb^2
vb = √2*g*H
vb = ( 2*9.81*15 ) ^0.5
vb = 17.15517 m/s
- The angular momentum of system before collision is:
M1 = ( 4.80 ) * ( 17.15517 ) * ( 6/2)
M1 = 247.034448 kgm^2 /s
- After collision, the momentum is transferred to the other ball. The momentum after collision is:
M2 = mo*vo*(L/2)
- From principle of conservation of angular momentum the initial and final angular momentum remains the same.
M1 = M2
vo = 247.03448 / (5.10*3)
vo = 16.14604 m/s
- The speed of the other ball after collision is (vo), the maximum height can be determined by using the principle of conservation of energy:
ΔP.E = ΔK.E
mo*g*h = 0.5*mo*vo^2
h = vo^2 / 2*g
h = 16.14604^2 / 2*(9.81)
h = 13.3 m
The conservation of energy and angular momentum allows to find the height of the ball of m = 5.1 kg after the collision in the system is:
- The height of the ball of m = 5.1 kg at the other end of the bar rises is: y = 14.4 m
Given parameters
- Mass of the ball, m₁ = 4.8 kg.
- Ball height, y₁ = 15.0 m
- Bar mass, M = 7.0 kg
- Bar length L = 6.0 m
- Mass at the other end of the bar m₂ = 5.10 kg
To find.
- The ball is hit in the collision, how much does the other end go up (ball 2)
Conservation of angular momentum establishes that if a system is isolated, the angular momentum is constant. Let's form the system by the three particles, in this case all the forces during the collision are internal and the angular momentum is conserved.
The angular momentum is defined by
L = r x p = r x (m v)
Where r is the distance and p is the linear moment, m the mass and v the velocity.
Let us write the angular momentum in two instants, with respect to the point of rotation.
Initial instant. Before the crash.
[tex]L_o = L_{ball 1} = m_1 v_1 \frac{L}{2}[/tex]
Final moment. After the crash.
[tex]L_f = L_{ball 2}+ L_{bar} \\L_f = m_1 v_2 \frac{L}{2} + I_{bar} w[/tex]
The moment of inertia of a bar rotating through its center is tabulated.
I = [tex]\frac{1}{12} M L^2[/tex]
The angular velocity of the bar is related to the linear velocity.
v = w r = [tex]w \ \frac{L}{2}[/tex]
[tex]w = \frac{2v}{L}[/tex]
Let's substitute.
[tex]L_f = m_2 v_2 \frac{L}{2} + \frac{1}{12} M L^2 ( \frac{2 \ v_2}{L} )[/tex]
[tex]L_f = \frac{L}{2} \ v_2 \ ( m_2 + \frac{M}{12}[/tex] )
The angular moemoto is preserved.
[tex]L_o = L_f \\m_1 v_1 \frac{L}{2} = \frac{L}{2} \ v_2 ( m_2 + \frac{M}{12} ) \\v_2 = \frac{m_1}{m_2+ \frac{M}{2} } \ v_1[/tex]
Let's use the conservation of energy to find the velocity of the ball as it hits the bar.
Starting point. Higher
Em₀ = U = m₁ g y₁
Lowest point. When he touches the bar.
Em_f = K = ½ m₁ v₁²
Energy is conserved.
Em₀ = Em_f
m₁ g y₁ = ½ m₁ v₁²
[tex]v_1 = \sqrt{2gy_1}[/tex]
We substitute.
[tex]v_2 = \frac{m_1}{m_2 + \frac{M}{12} } \ \sqrt{2 g y_1}[/tex]
Let us conclude
[tex]v_2 = \frac{4.8}{5.1 + \frac{7.0}{12} } } \ \sqrt{2 \ 9.8 \ 15.0}[/tex]
v₂ = 14.4 m
In conclusion using the conservation of energy and angular momentum we can find the height of the ball of m = 5.1 kg after the collision in the system is:
- The height of the ball of m = 5.1 kg at the other end of the bar rises is: y = 14.4 m
Learn more about angular momentum here: brainly.com/question/25677703
