What is the rebound speed of a bumper car after colliding with a wall?

In summary, a 450kg bumper car with a spring constant of 3x10^7 N/m collides with a solid wall at a speed of 2m/s, resulting in a maximum compression of 7.7mm. The question asks for the speed at which the car will rebound from the wall. The answer is 1.6m/s, but there seems to be some discrepancy and uncertainty due to the lack of information provided about potential energy and friction.
  • #1
futb0l
A 450kg bumper car, with a spring which have a spring constant of 3x10^7 N/m collides at a speed of 2m/s with a solid wall. It gives a maximum compression of 7.7mm. At what speed will the car rebound of the wall?

I am having trouble with this one... I don't know how to go about solving the problem. The answer btw is 1.6m/s
 
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  • #2
Do you know how to find the potential energy from the spring constant and compression? You should have a formula for that and a formula for kinetic energy as a function of speed. Once the car has rebounded, all that potential energy becomes kinetic energy.
 
  • #3
HallsofIvy said:
Do you know how to find the potential energy from the spring constant and compression? You should have a formula for that and a formula for kinetic energy as a function of speed. Once the car has rebounded, all that potential energy becomes kinetic energy.

Since the final speed is not equal to the initial speed isn't there some friction involved in this question??

DOes the question say anything about the surface upon which this car travels??
 
  • #4
stunner5000pt said:
Since the final speed is not equal to the initial speed isn't there some friction involved in this question??

DOes the question say anything about the surface upon which this car travels??

Nope, it doesn't say anything about the surface, it says its just a solid wall... which is why it is strange.
 
  • #5
futb0l said:
I am having trouble with this one... I don't know how to go about solving the problem. The answer btw is 1.6m/s

Are you sure about this ?

I get 1.9762 m/s

Use energy conservation.

KE (before collision) = PE (at max compression) + E (lost to friction)

Plugging in the numbers, you can find E (lost).

During the rebound, you will lose this same amount of energy to friction, so KE (final) = PE - E (lost)
 
  • #6
Gokul43201 said:
Are you sure about this ?

I get 1.9762 m/s
I agree. The car has 900 Joules of energy (.5mv^2=.5*450*2*2). A spring compression of .0077 m contains 899.35 joules of energy. If it loses another 10.65 joules in converting it back to kinetic energy, you end up with 878.7 J, so v^2=2*878.7/450; v=1.9762. A speed of 1.6 m. gives the car 576 J. The problem contains no information to account for such a loss of energy.

AM
 
  • #7
That (1.6 m)answer would require either a spring constant of about 2.5 * 10^7 N/m or a compression of about 7.0 mm or just a whole new set of numbers.
 
  • #8
Thanks, probably something wrong with this question, I am stunned myself.
 

FAQ: What is the rebound speed of a bumper car after colliding with a wall?

What is a collision?

A collision is an event where two or more objects come into contact with each other, exchanging energy and momentum. Collisions can be either elastic or inelastic, depending on whether or not there is a change in the kinetic energy of the objects involved.

What is Hooke's Law?

Hooke's Law is a principle in physics that describes the relationship between the force applied to an elastic object and the resulting deformation or change in length of the object. It states that the force applied is directly proportional to the amount of deformation, as long as the object remains within its elastic limit.

How is Hooke's Law related to collisions?

In collisions between elastic objects, Hooke's Law can be used to calculate the force exerted by each object on the other during the collision. This can help determine the resulting velocities and directions of the objects after the collision.

Can Hooke's Law be applied to all types of collisions?

No, Hooke's Law is only applicable to elastic collisions, where there is no loss of kinetic energy. In inelastic collisions, there is a change in kinetic energy and Hooke's Law cannot be used to accurately calculate the forces involved.

How is Hooke's Law used in real life?

Hooke's Law is used in a variety of real-life applications, such as designing springs for various mechanical devices, determining the strength of materials, and understanding the behavior of elastic materials like rubber and metal. It is also used in understanding the impact of collisions in sports, such as in the design of helmets and protective gear.

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