Elastic Collision of a ball and wire

In summary, in this conversation, Quincy asks for help in finding the tension in a wire after an elastic collision between a 10.0 kg ball hanging from the ceiling and a 1.20 kg ball moving horizontally at 7.50 m/s. The correct approach is to use conservation of both momentum and energy, taking into account the vertical acceleration of the small ball after the collision.
  • #1
Quincy
228
0

Homework Statement


An 10.0 kg ball, hanging from the ceiling by a light wire 125 cm long, is struck in an elastic collision by a 1.20 kg ball moving horizontally at 7.50 m/s just before the collision. Find the tension in the wire just after the collision.


Homework Equations


M1V1o + M2V1o = M1V2f + M2V2f

The Attempt at a Solution


(1.2 kg)(7.5 m/s) = (10 kg)(Vf)

Vf = 0.9 m/s

Fc = mV^2/R = (10 kg)(0.9^2)/(1.25 m) = 6.48 N

Fc = T - mg
6.48 = T - (10 kg)(9.8)

T = 104.48 N -- Book says it's incorrect, what am i doing wrong?
 
Last edited:
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  • #2
Hi Quincy! :smile:
Quincy said:
An 10.0 kg ball, hanging from the ceiling by a light wire 125 cm long, is struck in an elastic collision by a 1.20 kg ball moving horizontally at 7.50 m/s just before the collision. Find the tension in the wire just after the collision.

why will the tension be different if the ball has a horizontal velocity?

Because although it has a purely horizontal velocity, it has a vertical acceleration. :wink:

Use Ftotal = ma. :smile:
 
  • #3
tiny-tim said:
Hi Quincy! :smile:


Because although it has a purely horizontal velocity, it has a vertical acceleration. :wink:

Use Ftotal = ma. :smile:

I realized that right after i posted the problem :P, i edited it now but the answer's still incorrect.
 
  • #4
Quincy said:
(1.2 kg)(7.5 m/s) = (10 kg)(Vf)

Nooo … you've ignored the momentum of the small ball after the collision. :redface:

Try again, using conservation of both momentum and energy. :smile:
 

1. What is an elastic collision?

An elastic collision is a type of collision in which both kinetic energy and momentum are conserved. This means that the total energy and total momentum of the system before and after the collision are equal.

2. What is the difference between an elastic collision and an inelastic collision?

An elastic collision is a collision in which the objects involved bounce off each other without any loss of kinetic energy. In contrast, an inelastic collision is a collision in which some kinetic energy is lost, usually due to the objects sticking together or deforming upon impact.

3. How does the elasticity of the objects affect the outcome of a collision?

The elasticity of the objects involved in a collision determines whether the collision is elastic or inelastic. If the objects are perfectly elastic, they will bounce off each other without any loss of kinetic energy. If the objects are not perfectly elastic, some kinetic energy will be lost during the collision.

4. How is the speed of the objects after an elastic collision calculated?

The speed of the objects after an elastic collision can be calculated using the conservation of momentum and the conservation of kinetic energy equations. The equations are:
m1v1 + m2v2 = m1v1' + m2v2' (conservation of momentum)
1/2m1v1² + 1/2m2v2² = 1/2m1v1'² + 1/2m2v2'² (conservation of kinetic energy)

5. How can the elastic collision of a ball and wire be applied in real life?

The elastic collision of a ball and wire can be applied in real life in sports, such as billiards or pool, where the balls collide with each other and the walls of the table. It is also used in engineering and design, where the impact of objects is taken into account to ensure safety and efficiency. Additionally, it is used in physics experiments and simulations to understand the behavior of objects during collisions.

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