Tension in Wire: Solve 8kg Ball Elastic Collision

In summary, an 8kg ball hanging from a 1.35m string was hit by a 2kg ball moving horizontally at 5m/s in an elastic collision. The tension in the string just after the collision can be found by using conservation of momentum and conservation of energy equations. The final velocity of the 2kg ball is 1.25m/s and the total kinetic energy is conserved. The tension in the string must provide enough force to cancel out the weight of the hanging mass and supply the centripetal force.
  • #1
PsychonautQQ
784
10

Homework Statement


a 8kg ball is hanging from a 1.35m string. A 2 kg ball moving horizontally at 5m/s hits the hanging 8kg ball in an elastic collision. What is the tension in the string just after the collision?


Homework Equations





The Attempt at a Solution


Initially there is only a gravitational force of 8*9.8 = 78.4 N.

Using conservation of momentum, we can find that the ball has a velocity of 1.25 m/s just after the collision.

so the centripetal acceleration is v^2/r = 1.1574 m/s^2.
That centripetal acceleration is due to a centripetal force point in the opposite direction as the gravitation force

so F_g - M*a_c = answer
78.4 - (8)*(1.1574) =69.14 Newtons

The back of the book says the answer is 102 Newtons.

I feel like I worked this out right as far as my understanding of physics goes, but then I think about it and I feel like the tension should increase.

Even if I add the centripetal force to the gravitational force I come up short of 102 N.
Help?
 
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  • #2
What does it mean that the collision was elastic?

When you did the conservation of momentum calculation, what did you use for the final speed of the 2kg ball?
Looks like you guessed the ball comes to rest - how do you now it comes to rest?

That centripetal acceleration is due to a centripetal force point in the opposite direction as the gravitation force
So the tension in the string has to supply the centripetal force as well as enough to cancel out the weight of the hanging mass.
 
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  • #3
Elastic means no kinetic energy was lost in the collision. I guess the way I did it is not correct then. How do I find the final energy of the first mass? I thought in an elastic collision it would transfer all it's energy completely to the second mass.
 
  • #4
I thought in an elastic collision it would transfer all it's energy completely to the second mass.
That's not what it means... and it is also not what you did: check

EK (first mass) before collision: ##K_i= \frac{1}{2}(2\text{kg})(5\text{m/s})^2=25\text{ J}##
EK (transferred to second mass) after the collision: ##K_f= \frac{1}{2}(8\text{kg})(1.25\text{m/s})^2=6.25\text{ J}##

"Conserved" means it doesn't change at all - the total kinetic energy should be unchanged.

After the collision you have two unknown velocities.
You also have two equations - one for conservation of momentum, and the other for conservation of energy.
These equations have to hold simultaneously.
 
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  • #5


I understand your confusion and will try to provide an explanation for the discrepancy in the calculated tension and the answer provided in the book.

Firstly, your approach using conservation of momentum and centripetal acceleration is correct. However, there are a few factors that could affect the accuracy of your calculation.

One factor is the assumption that the collision is perfectly elastic. In reality, there will always be some energy lost to friction or other factors, which could affect the final velocity and therefore the tension in the string.

Another factor is the assumption that the string is massless. In reality, the string will have some mass and will also experience a force due to the collision, which could affect the tension in the string.

Additionally, the calculation of the centripetal force assumes that the string is taut and remains at a constant length during the collision. If the string were to stretch or become slack during the collision, this could also affect the tension in the string.

Lastly, it is possible that the answer provided in the book is incorrect. As a scientist, it is important to question and critically evaluate all sources of information, including textbooks and answer keys.

In conclusion, while your approach to solving the problem is correct, there are several factors that could affect the accuracy of your calculation. It is also important to consider and question the validity of the answer provided in the book. Further experimentation or theoretical analysis may be necessary to determine the exact tension in the string after the collision.
 

Related to Tension in Wire: Solve 8kg Ball Elastic Collision

1. What is tension in a wire and how does it affect a collision?

Tension in a wire is the force exerted by the wire when it is stretched or compressed. In the case of a collision, tension in the wire can affect the magnitude and direction of the force experienced by the objects involved.

2. How is tension in a wire calculated?

Tension in a wire can be calculated using the formula T = F * l, where T is tension, F is the force applied to the wire, and l is the length of the wire.

3. What is the difference between elastic and inelastic collisions?

Elastic collisions are collisions where the total kinetic energy of the system is conserved, meaning that the objects involved bounce off each other without any loss of energy. Inelastic collisions, on the other hand, involve a loss of kinetic energy due to deformation or other factors.

4. How does the mass of the objects involved affect the tension in the wire during a collision?

The mass of the objects involved can affect the tension in the wire during a collision by increasing or decreasing the force applied to the wire. Objects with greater mass will exert a greater force on the wire, resulting in higher tension.

5. What are some real-world applications of analyzing tension in a wire during collisions?

One real-world application of analyzing tension in a wire during collisions is in the design and testing of safety equipment, such as seatbelts and airbags, which must withstand the force of a collision without breaking. Another application is in sports, where understanding the tension in a wire can help improve the performance and safety of equipment, such as tennis rackets or rock climbing ropes.

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