Elastic head-on collision problem

In summary, the way I learned to solve this was to switch to a frame of reference where one object is stationary.
  • #1
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Homework Statement
In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The collision is cushioned by a spring (k = 1200 N/m).
1 Determine the velocity of each cart after the collision.
2 Determine the maximum compression of the spring.
Relevant Equations
n/a
The way I learned to solve this was to switch to a frame of reference where one object is stationary.

given: m1 =0.6kg v1 = 5.0m/s [W], m2 = 0.8kg v2 = 2.0 m/s [E]

Setting v2 to rest by adding 2.0 m/s W to each object

New velocities are v1 = 7.0 m/s [E] and v2 = 0.0m/s

Then using the equations for v1

v1f = V1 (m1 - m2) / (m2 + m1)

= (7.0)(0.6-0.8) / (0.8-0.6)

= -7 m/sand for v2f

v2f = 2m1(v1) / (m2+m1)

= 2 (0.6) (7.0) / (0.8 + 0.6)
= 6 m/s

Switching back to Earth's frame of reference

v1f = 1.8 m/s
v2f = 4.0 m/s

I get these answers, yet i see other answers online using conservation of energy and they get different answers from me. Am I doing this properly?
 
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  • #2
I haven't checked your answer, as you can do it yourself:

Is momentum conserved (in your answer)? Is energy conserved (in your answer)?
 
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  • #3
If you have one object moving at 2.0 m/s and another at 5.0 m/s, you need to add (-2 m/s) to each object to get the 2 m/s object to be at rest. Thus, in that frame the 5 m/s object will have a velocity of 3 m/s.
 
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  • #4
kuruman said:
If you have one object moving at 2.0 m/s and another at 5.0 m/s, you need to add (-2 m/s) to each object to get the 2 m/s object to be at rest. Thus, in that frame the 5 m/s object will have a velocity of 3 m/s.

This is what I did in my first attempt but in my lesson it shows an example like this:
1615919848179.png

Which is why in my attempt above I added the 2.0m/s onto the 5.0 m/s. Am I misunderstanding this?
 
  • #5
Look at the sentence in bold letters: Always add the opposite to the initial velocity of object 2. In the second example object 2 has velocity -4 m/s so you add the opposite of that which is +4 m/s. In your example object 2 has velocity +2 m/s so you add the opposite of that which is ##\dots##?
 
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  • #6
kuruman said:
Look at the sentence in bold letters: Always add the opposite to the initial velocity of object 2. In the second example object 2 has velocity -4 m/s so you add the opposite of that which is +4 m/s. In your example object 2 has velocity +2 m/s so you add the opposite of that which is ##\dots##?
Alright thank you.

This is what I've done now:

Let West be the positive direction

v1f = V1 (m1 - m2) / (m2 + m1)

= (3.0)(0.6-0.8) / (0.8-0.6)
= -0.44 m/s

v2f = 2m1(v1) / (m2+m1)

= 2 (0.6) (3.0) / (0.8 + 0.6)
= 2.57 m/s

So if this is done properly the answers would be v1 = 2.44 m/s [W] and v2 = -4.57 m/s [E] ??
 
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  • #7
As suggested by @PeroK, if this is done properly then the momentum conservation equation
Pbefore = Pafter should hold. If I put in your numbers,
Pbefore = 0.6*5 + 0.8*2 and Pafter = 0.6*(2.44) + 0.8*(4.57)
Does the equality hold? This is a test you can do yourself.
 
  • #8
Traced said:
Alright thank you.

This is what I've done now:

Let West be the positive direction

v1f = V1 (m1 - m2) / (m2 + m1)

= (3.0)(0.6-0.8) / (0.8-0.6)
= -0.44 m/s

v2f = 2m1(v1) / (m2+m1)

= 2 (0.6) (3.0) / (0.8 + 0.6)
= 2.57 m/s

So if this is done properly the answers would be v1 = 2.44 m/s [W] and v2 = -4.57 m/s [E] ??
You need a system here! There's no hope using E and W other than at the beginning and end. The data is:

##m_1 = 0.6 kg##, ##u_1 = 5.0 m/s##, ##m_2 = 0.8 kg##, ##u_2 = - 2.0 m/s##

To get the rest frame of ##m_2## you add ##2 m/s##; to go back to the original frame you add ##-2 m/s##.

PS If W is positive, then it's either ##2 m/s## East or ##-2.0 m/s##, but not ##-2 m/s## East. That's too confusing. One or the other, please!
 
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  • #9
Alternate Method.

Before I present this I would advise PF user Traced that this method involves the use of a formula which one would need (in principle) to derive for oneself before using. The formula is derived using the same well established principles of conservation of momentum and energy which are routinely used for solving collision problems. But - analogous to completing the square in quadratic problems - you can repeatedly complete the square or you can simply use the general formula that derives from completing the square for the general quadratic ##ax^2+bx+c=0## in the first instance.

In a collision the forces acting on the colliding bodies are equal and opposite. Since the "impact time" for both bodies is the same, it follows that they experience equal and opposite impulses. So what we can do is simply calculate that impulse (that's where the formula comes in) and apply it as an equal and opposite change in momentum. Thereby calculate the post-collision momentum for both and divide by their masses to obtain post-collision velocities.

For an elastic collision: $$\Delta P=2\mu \Delta v$$ where ##\mu## is the 'reduced mass' of the colliding bodies: $$\mu=\frac{m_1 m_2}{m_1+m_2}$$ and ##\Delta v## is their relative velocity.
 
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1. What is an elastic head-on collision?

An elastic head-on collision is a type of collision between two objects where both objects are moving towards each other and collide without any external forces acting on them. In this type of collision, the total kinetic energy of the system is conserved.

2. How is the velocity of the objects calculated in an elastic head-on collision?

The velocity of the objects after an elastic head-on collision can be calculated using the conservation of momentum and conservation of kinetic energy equations. The final velocities can be determined by solving these equations simultaneously.

3. What factors affect the outcome of an elastic head-on collision?

The outcome of an elastic head-on collision is affected by factors such as the masses of the objects, their initial velocities, and the coefficient of restitution (a measure of how much energy is lost during the collision). Other factors, such as the elasticity of the objects and the angle of collision, can also play a role.

4. Can an elastic head-on collision result in a total loss of kinetic energy?

No, an elastic head-on collision is defined as a collision where the total kinetic energy of the system is conserved. This means that the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Therefore, there can never be a total loss of kinetic energy in an elastic head-on collision.

5. How is an elastic head-on collision different from an inelastic collision?

In an inelastic collision, some of the kinetic energy is lost during the collision and converted into other forms of energy, such as heat or sound. This means that the total kinetic energy of the system is not conserved. In contrast, an elastic head-on collision is defined by the conservation of kinetic energy, where no energy is lost during the collision.

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