Solving for Elastic Collision b/w Thomas & Diesel

[stunner5000pt]

Homework Statement

Thomas the tank engine and diese are involved in an elastic collision. Thomas = 2.5kg and Diesel = 5.0kg . Thomas is initially at rest and diesel is at 0.60 m/s. The force separation graph for the ensuing collision is given below:

a. What is the total kinetic energy before the collision? After?
b. What is the velocity of each train at minimum separation?
c. What is the total kinetic energy at minimum separation?
d. How much energy is stored at minimum separation?
e. What is the magnitude of the force acting on each mass at minimum separation?
f. What is the minimum separation distance between the trains? Hint: The energy temporarily stored at minimum separation equals a portion of the area under the above graph. The collision starts when the centers of the trains are separated by 0.03 m as shown on the above graph at which time the collision force is 15 N. But this force increases to 30 N and then eventually 45 N.

Homework Equations

$$E_{K} = \frac{1}{2} mv^2$$
$$\Delta p = m \Delta v = F \Delta t$$

The Attempt at a Solution

A. What is the total kinetic energy before the collision? After?
$$E_{K} = \frac{1}{2} 5 (0.6)^2 = 0.9 J$$

B.What is the velocity of each train at minimum separation?
Don't we need to calculate minimum separation before we compute this? Do we assume that at minimum separation that this in an inelastic collision? In which case we should make use of this formula :

$$m_{D} v_{D} = \left( m_{T} + m_{D} \right) v$$
and solve for v?
v = 0.4 m/s

C. What is the total kinetic energy at minimum separation?
Use the value of v calculated from b above in $$\frac{1}{2} \left( m_{T} + m_{D} \right) v^2 = 0.6 J$$

D. How much energy is stored at minimum separation?
Energy stored as minimum separation. Do we simply subtract the kinetic energy calculated in A from the energy calculated in C?
Would that be 0.3 J?

E. What is the magnitude of the force acting on each mass at minimum separation?
At minimum separation, as there is 0.3 J, the force acting would be 30J

F. What is the magnitude of the force acting on each mass at minimum separation?
The 0.3 J is represented as part of the area under the graph. It requires 0.15 J to compress the system between the train 0.01m and the additional 0.15J would compress the system an additional 0.005m. This would result in a minimum separation distance of 0.015 m

Please let me know if this is all correct? Thanks for your help.

 

[anathema]

Your analysis is mostly correct, but here are some clarifications and corrections:

A. Your calculation for the initial kinetic energy of Diesel is correct, but you also need to calculate the initial kinetic energy of Thomas, which is 0 since it is at rest.

B. The velocity of each train at minimum separation can be calculated using conservation of momentum, as you mentioned. However, since this is an elastic collision, the total kinetic energy should be conserved as well. This means that the final kinetic energy should also be 0.9 J, and you can use this to solve for the final velocities of both trains.

C. Your calculation for the total kinetic energy at minimum separation is correct.

D. The energy stored at minimum separation is the difference between the initial and final kinetic energies, which is also equal to the area under the graph. So it would be 0.9 J - 0.6 J = 0.3 J.

E. The force acting on each mass at minimum separation is not 30 J, but rather 30 N. Remember that force is measured in Newtons, not Joules.

F. Your calculation for the minimum separation distance is incorrect. The area under the graph represents the total energy stored, not just the energy stored at minimum separation. To find the minimum separation distance, you need to find the point on the graph where the force is 30 N and calculate the corresponding distance. This can be done by finding the area of the triangle formed by the points (0,0), (0.01,15), and (x,30) and setting it equal to 0.3 J. You should get a minimum separation distance of approximately 0.02 m.

Overall, your analysis is correct but requires some minor corrections. Keep up the good work!