# Collision Question dealing with Spring Compression

• chubbyorphan
In summary: Therefore, the spring compresses a maximum of 0.12m at a spring constant of 1200 N/mAnd yes, you are correct, the answer should be sqrt(0.014) = 0.12m. Good catch!
chubbyorphan

## Homework Statement

Forum! I'm struggling on this one D:
any help would be greatly appreciated!

An elastic head-on collision, a 0.60kg cart moving 5.0m/s [W] collides with a 0.8-kg cart moving 2.0 m/s[E]
The collision is cushioned by a spring (k=1200N/m)

**Find the maximum compression of the spring
I'm not sure if I did this right..^

## The Attempt at a Solution

** note that v prime (v’) is final velocity

Required: maximum compression of the spring
Given: k = 1200 N/m
Analysis and Solution:
Let [W] represent positive
First we must find how fast the carts are moving at the point where the spring is at maximum compression. At this point, the two carts must have the same velocity; otherwise the spring would not be compressed to a maximum. To find the velocity, use conservation of momentum:

PT = PT’
m1v1 + m2v2 = (m1 + m2)v’
(0.60)(5.0) + (0.80)( –2.0) = (0.60 + 0.80) v’

Notice that v2 is negative because it is traveling [E] and we already decided [W] is positive

(0.60)(5.0) + (0.80)( –2.0) = (0.60 + 0.80) v’
3 + (–1.6) = (1.40) v’
1.4 = (1.40) v’
v’ = 1.0m/s
Therefore, at maximum compression, the two carts are moving at 1.0m/s
****^this answer feels funny to me..

Now use conservation of momentum to find the maximum compression of the spring:

(1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)(m1 + m2)v’^2 + (1/2)kx << could someone please explain this out in words how my equation makes sense.. Its from my book and looking at it is confusing me
Recall k = 1200N/m
(1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)(m1 + m2)v’^2 + (1/2)(1200N/m)x

Solve for x (the maximum compression of the spring)

(1/2)(0.60)(5.0)^2 + (1/2)(0.80)(2.0)^2 = (1/2)(0.60 + 0.80)(1.0)^2 + (1/2)(1200)x

Notice that v2 is now positive despite the fact that it represents 2.0m/s [E].. and our forward direction was already decided to be [W]

7.5 + 1.6 = 0.70 + 600x
600x = 8.4
x = 0.014
Therefore, the spring compresses a maximum of 1.4cm or 0.014m at a spring constant of 1200 N/m
...^is this right :S

back to the purple text.. I understand why I would make the 2.0m/s[E] negative by the first purple text..but then in the second part of my solution(where the second purple text is).. as per following the along with the book.. it instructs me to suddenly make this value positive..
Can someone please explain why this makes sense D:

Last edited:
chubbyorphan said:

## Homework Statement

Forum! I'm struggling on this one D:
any help would be greatly appreciated!

An elastic head-on collision, a 0.60kg cart moving 5.0m/s [W] collides with a 0.8-kg cart moving 2.0 m/s[E]
The collision is cushioned by a spring (k=1200N/m)

**Find the maximum compression of the spring
I'm not sure if I did this right..^

## The Attempt at a Solution

** note that v prime (v’) is final velocity

Required: maximum compression of the spring
Given: k = 1200 N/m
Analysis and Solution:
Let [W] represent positive
First we must find how fast the carts are moving at the point where the spring is at maximum compression. At this point, the two carts must have the same velocity; otherwise the spring would not be compressed to a maximum. To find the velocity, use conservation of momentum:

PT = PT’
m1v1 + m2v2 = (m1 + m2)v’
(0.60)(5.0) + (0.80)( –2.0) = (0.60 + 0.80) v’

Notice that v2 is negative because it is traveling [E] and we already decided [W] is positive

(0.60)(5.0) + (0.80)( –2.0) = (0.60 + 0.80) v’
3 + (–1.6) = (1.40) v’
1.4 = (1.40) v’
v’ = 1.0m/s
Therefore, at maximum compression, the two carts are moving at 1.0m/s
****^this answer feels funny to me..
Looks very reasonable. It also happens to be the speed of the center of mass of the system.
Now use conservation of momentum to find the maximum compression of the spring:

(1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)(m1 + m2)v’^2 + (1/2)kx << could someone please explain this out in words how my equation makes sense.. Its from my book and looking at it is confusing me
It's saying that the kinetic energy before collision is equal to the kinetic energy when the masses are moving as one (same velocity), plus the energy stored in the spring. It's just a statement of conservation of energy.
Recall k = 1200N/m
(1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)(m1 + m2)v’^2 + (1/2)(1200N/m)x

Solve for x (the maximum compression of the spring)

(1/2)(0.60)(5.0)^2 + (1/2)(0.80)(2.0)^2 = (1/2)(0.60 + 0.80)(1.0)^2 + (1/2)(1200)x
The energy of a spring is given by (1/2)kΔx2. Don't forget that the Δx is squared!

Notice that v2 is now positive despite the fact that it represents 2.0m/s [E].. and our forward direction was already decided to be [W]

7.5 + 1.6 = 0.70 + 600x
600x = 8.4
x = 0.014
Therefore, the spring compresses a maximum of 1.4cm or 0.014m at a spring constant of 1200 N/m
...^is this right :S

back to the purple text.. I understand why I would make the 2.0m/s[E] negative by the first purple text..but then in the second part of my solution(where the second purple text is).. as per following the along with the book.. it instructs me to suddenly make this value positive..
Can someone please explain why this makes sense D:
[/QUOTE]
Squaring any value (except for imaginaries, of course) results in a positive result. So the sign of the velocities doesn't matter when you're working with kinetic energy.

thanks so much gneill! That thread was up all day and you were the only one willing to take the time to the read it >.>

and thanks for catching my mistake! I haven't handed it in yet thankfully!

so my answer should be √0.014
rounding to two sig digits:
x = 0.12

## 1. What is spring compression?

Spring compression is the change in length or deformation of a spring when a force is applied to it. This force causes the coils of the spring to compress and store potential energy.

## 2. How is spring compression related to collision?

In collisions, objects come into contact and exert forces on each other. When a spring is involved in a collision, the force of the collision causes the spring to compress. This compression can be used to analyze and understand the collision.

## 3. What factors affect the amount of spring compression in a collision?

The amount of spring compression in a collision is affected by the mass and velocity of the objects involved, as well as the stiffness of the spring. The greater the mass and velocity, and the stiffer the spring, the more the spring will compress.

## 4. How can spring compression be calculated?

The amount of spring compression can be calculated using Hooke's Law, which states that the force applied to a spring is directly proportional to the amount of compression or extension. This can be represented by the equation F = -kx, where F is the force, k is the spring constant, and x is the displacement or compression of the spring.

## 5. What is the significance of spring compression in collisions?

Spring compression is important in collisions because it allows for the analysis of the forces involved and the energy transferred during the collision. This information can be used to understand the outcome of the collision and make predictions about future collisions.

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