Question

A glider of mass m1 = 10.0 kg initially moving towards the right at a
speed of v1i = 2.00 m/s on a frictionless airtrack collides with a
spring attached to a second glider of mass m2 = 4.0 kg initially
moving toward the left with a speed of v2i = 5.00 m/s. The spring
constant is 700 N/m.
(a) During the collision, there is an instant where both gliders briefly come to rest ( v1i = v2i = 0 m/s). At
this instant, how much is the spring compressed? (Hint: the spring is a conservative force –
conservation of mechanical energy is appropriate)
(b) Because the spring force is conservative, all the energy put into it during the collision will be put
back into the gliders, i.e. total kinetic energy before and after will be conserved. What is the final
velocity (speed and direction) of both gliders?


In the first part the sum of the KEnergies is converted to elastic potential energy. The elastic potential energy is equal to the sum of the KE.

For the second part, you have two fundamental principles: The KE is conserved (sum of initial KE is equal to the final KE, and the momentum is conserved). This will be enough to solve. The algebra may be tricky.


I understand how to do the second part now, but im still lost on how to setup the first part. Please help!


First Part

1/2 m1 * 2^2 + 1/2 m2 * 5^2= 1/2 k x^2