Energy Conservation and Coefficient of Restitution

[B3NR4Y]

Homework Statement

A 1300-kg car is backing out of a parking space at 5.0 m/s . The unobservant driver of a 1700-kg pickup truck is coasting through the parking lot at a speed of 3.2 m/s and runs straight into the rear bumper of the car.

(a). What is the change in internal energy of the two-vehicle system if the velocity of the pickup is 1.5 m/s backward after they collide?
(b). What is the coefficient of restitution for this collision?

Homework Equations

[itex] p_{total, \, initial} = p_{total, \, final} = m_{c} v_{ci} + m_{p} v_{pi} = m_{c} v_{cf} + m_{p} v_{pf} [/itex]
Conservation of momentum, you can use it to solve for the velocity of the car after the collision.
[itex] e = -\frac{v_{12, \, f}}{v_{12, \, i}} [/itex]
where v₁₂ is the car's velocities relative to each other. e is the coefficient of restitution.
[itex] K = \frac{1}{2} mv^{2} [/itex]
[itex] \Delta K = -\Delta E [/itex]

The Attempt at a Solution

To start the problem, I drew a diagram and defined my coordinate system such that the initial velocity of the pickup is in the positive direction. Since the author of this textbook ignores two dimensional motion for 10 chapters, the collision is one-dimensional therefore the truck drives directly into the rear of the car. So, the initial velocity of the car is -5.0 m/s, because if is coming toward the pickup. The pickup is going backward after the collision, therefore having negative velocity. So I set up my momentum equation,
[itex] (1300 kg)(-5.0 m/s) + (1700 kg)(3.2 m/s) = (1300 kg)v_{c, \, f} + (1700 kg)(-1.5 m/s) [/itex]
I got the initial total momentum to be -1060 and used algebra (adding 2550 kg m/s to both sides and dividing by 1,300) to arrive at my answer of ~2.78 m/s for the velocity of the car. Using my calculator I confirmed it agreed with the conservation of momentum (same magnitude, just different directions). I started the problem with the easiest question, which was (b). and got 0.52, incorrect. I set up my problem:

[itex] e = \frac{2.78+1.5 (m/s)}{8.2 (m/s)} [/itex]

The denominator is flipped because I distributed the negative sign, but got the same answer if I didn't so it doesn't matter much. And I used the real, not rounded, value for the velocity of the car so that is not the source of my error.

 

[haruspex]

[itex] e = \frac{2.78+1.5 (m/s)}{8.2 (m/s)} [/itex]

If the posterior velocities are -2.78 and -1.5, what is the posterior relative velocity?

 

[B3NR4Y]

If the posterior velocities are -2.78 and -1.5, what is the posterior relative velocity?

~-1.25 m/s, I would say

 

[haruspex]

~-1.25 m/s, I would say

Well, 1.28, not caring about the sign for now; not 2.78+1.5, as in your original post.

 

[HalfLostAndSearching]

I would like to commend you on your thorough and detailed approach to solving this problem. It is clear that you have a good understanding of conservation of momentum and the concept of coefficient of restitution.

In order to solve for the coefficient of restitution, we need to use the equation:

e = - v12f / v12i

where v12f is the final relative velocity of the two vehicles and v12i is the initial relative velocity. In this case, the initial relative velocity is (5.0 m/s + 3.2 m/s) = 8.2 m/s, since they are moving in opposite directions.

Using the value you calculated for the final velocity of the car (2.78 m/s) and the given final velocity of the pickup (-1.5 m/s), we get:

e = -(-1.5 m/s) / (8.2 m/s) = 0.183

This is the correct value for the coefficient of restitution. It is important to note that the negative sign in front of the initial relative velocity is already taken into account when we use the final velocities.

Therefore, the change in internal energy of the two-vehicle system is:

ΔE = -ΔK = - (1/2)(1300 kg)(2.78 m/s)^2 - (1/2)(1700 kg)(-1.5 m/s)^2 = - 6066.85 J

This negative value indicates that energy was lost during the collision, which is expected in a real-world scenario.

Overall, your approach and calculations were correct. The only error was in the final step of calculating the coefficient of restitution, where you mistakenly added the two final velocities instead of subtracting them. Keep up the good work!