Springs and Elastic Coollisions

[fibb]

Homework Statement

A ball of mass 2.5kg is moving to the left along a smooth, horizontal table at 3.5m/s. An idea spring (one that obeys Hooke's Law) with a spring contstant of k=1100N/m, and a relaxed length of 0.3m, is attached to its front. It collides head on with a 4.5 kg ball, initially at rest.
a) What is the velocity of each ball at minimum separation?
b) What is the change in total kinetic energy at minimum separation? as compared to initial K.E. of system?

Homework Equations

p=mv
Hooke's Law: Fx=kx
Ee=1/2kx^2
I know there's more equations needed

The Attempt at a Solution

For a, I found an expression of v1 and v2 with the momentum formula and conservation of momentum and tried to eliminate. But I got stuck at that part.
I don't really know how to do b.

 

[physixguru]

http://hyperphysics.phy-astr.gsu.edu/Hbase/elacol.html

 

[Moetasim]

I would first like to clarify that this type of problem falls under the field of classical mechanics and specifically, the study of collisions and energy conservation. With that in mind, let's begin with the solution to the given problem.

a) To find the velocity of each ball at minimum separation, we can use the principle of conservation of momentum. This means that the initial momentum of the system (ball 1 + ball 2) must be equal to the final momentum at minimum separation. We can express this as:

m1v1 + m2v2 = (m1+m2)vf

Where m1 and m2 are the masses of the two balls, v1 and v2 are their initial velocities, and vf is their final velocity at minimum separation. Plugging in the given values, we get:

(2.5 kg)(3.5 m/s) + (4.5 kg)(0 m/s) = (2.5 kg + 4.5 kg)vf

Solving for vf, we get vf = 1.75 m/s. This means that both balls will have a velocity of 1.75 m/s at minimum separation.

b) To find the change in total kinetic energy at minimum separation, we can use the principle of conservation of energy. The total kinetic energy of the system is equal to the sum of the kinetic energy of each ball. We can express this as:

KEtotal = KE1 + KE2

Where KE1 and KE2 are the kinetic energies of ball 1 and ball 2, respectively. Using the equation for kinetic energy, KE = 1/2mv^2, we can calculate the initial kinetic energy of the system as:

KEinitial = (1/2)(2.5 kg)(3.5 m/s)^2 + (1/2)(4.5 kg)(0 m/s)^2 = 15.625 J

At minimum separation, the final kinetic energy of the system can be calculated as:

KEfinal = (1/2)(2.5 kg)(1.75 m/s)^2 + (1/2)(4.5 kg)(1.75 m/s)^2 = 10.9375 J

Therefore, the change in total kinetic energy at minimum separation is:

ΔKE = KEfinal - KEinitial = 10.9375 J - 15.625 J = -4.687