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Answer :
The tension in the wire just after an 8.00 kg ball is struck in an elastic collision by a 2.00 kg ball involves determining the gravitational force and the change in horizontal momentum due to the collision. To calculate this tension, we would need additional information about the ball's motion post-collision.
The question relates to the topic of elastic collisions and physics concepts involving tension in wires and motion. Since the wire’s length and the mass of the ball are provided, you would start by finding the forces acting on the ball just after the collision. The tension in the wire can be found by combining the force due to gravity and the horizontal component of the ball’s momentum change due to the collision.
To solve this problem, we need to use the principle of conservation of momentum and energy.
Before the collision, the momentum of the system is:
p = m1v1 + m2v2
where m1 = 8.00 kg is the mass of the hanging ball, v1 = 0 (since it is at rest), m2 = 2.00 kg is the mass of the moving ball, and v2 = 5.00 m/s is its velocity. Therefore, the initial momentum of the system is:
p_initial = m1v1 + m2v2 = 2.005.00 = 10.00 kgm/s
After the collision, the 2.00 kg ball will stick to the 8.00 kg ball, and they will move together as one body. Since the collision is elastic, the total mechanical energy of the system is conserved. The mechanical energy of the system before the collision is:
E_initial = (1/2)m1v1² + (1/2)m2v2²= 0.52.005.00^2 = 25.00 J
The mechanical energy of the system after the collision is:
E_final = (1/2)MV²
where M = m1 + m2 = 10.00 kg is the mass of the combined system, and V is the velocity of the combined system just after the collision.
Using the principle of conservation of momentum, we know that:
p_initial = p_final
or
m1v1 + m2v2 = (m1 + m2)*V
Substituting the values we know, we get:
8.000 + 2.005.00 = (8.00 + 2.00)*V
V = 1.00 m/s
So, the velocity of the combined system just after the collision is 1.00 m/s.
Now, we can calculate the mechanical energy of the system after the collision:
E_final = (1/2)MV^2 = 0.510.001.00²= 5.00 J
Since the total mechanical energy of the system is conserved, we have:
E_final = E_initial
Therefore, the kinetic energy lost during the collision is:
ΔK = E_initial - E_final = 25.00 - 5.00 = 20.00 J
This kinetic energy is dissipated in the form of internal energy, such as heat, sound, and deformation of the balls.
Finally, we can find the tension in the wire just after the collision by considering the forces acting on the combined system. Since the system is in equilibrium, the tension in the wire must be equal to the weight of the system:
Tension = Weight = M*g
where g = 9.81 m/s² is the acceleration due to gravity.
Substituting the values we know, we get:
Tension = 10.00*9.81 = 98.1 N
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Rewritten by : Barada
The tension in the wire just after the collision is 102.18 N.
The conservation of momentum and energy to determine the velocities of the two balls just after the collision. After that, we will find the tension in the wire.
Momentum Conservation:
[tex]\[ m_1 \cdot u_1 + m_2 \cdot u_2 = m_1 \cdot v_1 + m_2 \cdot v_2 \][/tex]
Kinetic Energy Conservation (Elastic Collision):
[tex]\[ \frac{1}{2} m_2 \cdot u_2^2 = \frac{1}{2} m_1 \cdot v_1^2 + \frac{1}{2} m_2 \cdot v_2^2 \][/tex]
The conservation of momentum:
[tex]\[8.00 \cdot 0 + 2.00 \cdot 5.00 = 8.00 \cdot v_1 + 2.00 \cdot v_2\][/tex]
[tex]\[10 = 8v_1 + 2v_2 \quad \text{(Equation 1)}\][/tex]
From conservation of energy:
[tex]\[\frac{1}{2} \cdot 2.00 \cdot 5.00^2 = \frac{1}{2} \cdot 8.00 \cdot v_1^2 + \frac{1}{2} \cdot 2.00 \cdot v_2^2\][/tex]
[tex]\[25 = 4v_1^2 + v_2^2 \quad \text{(Equation 2)}\][/tex]
Solve for [tex]\(v_1\) and \(v_2\)[/tex]
Equation 1 for one variable:
[tex]\[v_2 = 5 - 4v_1 \quad \text{(substituting in Equation 2)}\][/tex]
[tex]\[25 = 4v_1^2 + (5 - 4v_1)^2\][/tex]
[tex]\[25 = 4v_1^2 + 25 - 40v_1 + 16v_1^2\][/tex]
[tex]\[20v_1^2 - 40v_1 = 0\][/tex]
[tex]\[v_1(20v_1 - 40) = 0\][/tex]
[tex]\[v_1 = 0 \quad \text{or} \quad v_1 = 2\][/tex]
For [tex]\(v_1 = 2\)[/tex]:
[tex]\[v_2 = 5 - 4 \cdot 2 = -3\][/tex]
The Tension in the Wire
For [tex]\(v_1 = 2 \, \text{m/s}\)[/tex], use the formula for tension considering the centripetal force:
[tex]\[T = m_1 \cdot g + \frac{m_1 \cdot v_1^2}{R}\][/tex]
where [tex]\(g = 9.81 \, \text{m/s}^2\) and \(R = 1.35 \, \text{m}\)[/tex]:
[tex]\[T = 8.00 \cdot 9.81 + \frac{8.00 \cdot 2^2}{1.35}\][/tex]
[tex]\[T = 78.48 + \frac{32}{1.35} \approx 78.48 + 23.7 = 102.18 \, \text{N}\][/tex]
