How Do You Calculate Spring Compression and Cart Speed After Collision?

[x86]

Homework Statement

A 1.2 kg dynamics cart is rolling to the right along a horizontal lab desk at 3.6 m/s, when it collides head on with a spring bumper that has a spring constant of 2.00 * 10^2 N/m.

a) Determine the maximum compression of the spring
b) Determine the speed of the cart at the moment that the spring was compressed by 0.10 m.

Homework Equations

E mechanical = E potential + E kinetic
E kinetic = 1/2 m v^2
E elastic = 1/2 k x^2
E gravitational = mgh

The Attempt at a Solution

a) I got 0.28 m, and the book agrees.

b) The answer should be 1.3 m/s. Here's what I get:

Eg is constant, because the height doesn't change. Equal is:

Et = Ee + Ek

In the previous solution, i set Ee2 + 0 = Ek1 + 0 since before there is no elastic energy and afterwards there is no kinetic energy

So Et = 1/2 * 2 * 10 ^2 N/m * 0.28 ^ 2 m ^2 = 7.84 joules

Et - Ee = Ek

7.84 J - 1/2 * 2 * 10^2 N/m * 0.10^2 m^2 = Ek = 6.84 J

1/2 * 1.2 kg * v^2 = 6.84 J

sqrt(6.84 / 1.2 kg / 0.5) = v^2 = 11.4 m^2 / s^2

Square rooted I get 3.4 m/s

The book says I'm doing b wrong, but I don't understand what I did wrong, or if the book is wrong.

 

[Sunil Simha]

I'm getting the same answer as you.

 

[sankalpmittal]

Square rooted I get 3.4 m/s

The book says I'm doing b wrong, but I don't understand what I did wrong, or if the book is wrong.

Hello !

Do not worry. You are correct and if book does not concur with you, its wrong.

I get 3.36 m/s which is approximately same as your answer.

 

[haruspex]

Fwiw, you can get the book answer by making the mistake of setting the KE equal to the PE when the spring is compressed .1 m.

 

[Nickie]

Dear student,

Your approach to solving the problem is correct. However, there seems to be a mistake in your calculation for the final speed of the cart. The correct value should be 1.3 m/s, as the book states.

Let's go through your calculation step by step:

1. You correctly determined the maximum compression of the spring to be 0.28 m.

2. You then correctly set up the equation Et = Ee + Ek, where Et represents the total mechanical energy, Ee represents the elastic potential energy, and Ek represents the kinetic energy.

3. You correctly stated that the gravitational potential energy is constant throughout the problem, as the height does not change.

4. You correctly set up the equation Et = 1/2 * k * x^2 to calculate the total mechanical energy, where k is the spring constant and x is the maximum compression of the spring.

5. You correctly calculated the total mechanical energy to be 7.84 J.

6. You then correctly set up the equation Et - Ee = Ek to calculate the final kinetic energy of the cart.

7. However, there seems to be a mistake in your calculation of the final kinetic energy. It should be:

Et - Ee = Ek

7.84 J - 1/2 * 2 * 10^2 N/m * (0.28 m)^2 = Ek

7.84 J - 7.84 J = Ek

Ek = 0 J

This means that all of the initial mechanical energy (7.84 J) is converted into elastic potential energy when the spring is compressed by 0.28 m. Therefore, at this point, the cart has come to a complete stop and has no kinetic energy.

8. To calculate the speed of the cart at the moment when the spring is compressed by 0.10 m, we can use the equation Ek = 1/2 * m * v^2.

Ek = 1/2 * m * v^2

0 J = 1/2 * 1.2 kg * v^2

v^2 = 0 m^2/s^2

v = 0 m/s

Therefore, at this point, the cart has come to a complete stop and has no speed.

9. To calculate the final speed of the cart, we can use the equation v^2 = u^2 + 2as, where u is