Posted by Nick on Sunday, March 25, 2012 at 7:28pm.
Tom the train and Dan are involved in an elastic collision. A 2.5 kg Tom is, at rest but is approached headon by a 5.0 kg Dan moving at 0.60 m/s. The forceseparation graph for the ensuing collision is given:
the picture of the graph is found here: i54.tinypic (dot) com/2zyftae.jpg
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 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.
f. What is the magnitude of the force acting on each mass at minimum separation?
I really have no idea how to get started with this question, i know for instance that there are two basic types of collisions, elastic and inelastic. This one is an elastic one so you would use the ((m1m2)/(m1m2))v1 and/or ((2m1)/(m1+m2))v1.

Physics  bobpursley, Sunday, March 25, 2012 at 8:13pm
If you have no idea, you are in trouble.
a. KE before equals sume of the KE of both (and one is zero).
b. Elastic: KE after is same as before.
c. KE at min separation? zero KE, movement is stopped.
d. Now, the hard part. The area under the graph, in each part, is equal to energy (force*distance). When the area is equal to twice that before KE, that separation is min distance.
e. above
f. read the graph.

Physics  Triple, Tuesday, December 16, 2014 at 9:43pm
@bobpursley
KE is not zero at minimum separation. Key word is minimum separation.
For b at minimum separation the velocities of both objects is equal. Both masses are also treated as one (add them both).
Also momentum is conserved even though Ek temporarily drops at minimum separation.
Pbefore collision = P@ min separation
Use P=mv to find the velocity at minimum separation.
(m1*v1) = (m1+m2)*V@min separation
Solve for V.
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