Physics perfectly elastic collision problem with pucks?

Question

I can't seem to figure this one out after using the equations my teacher gave me. I don't know what I'm doing wrong. Please help!


A blue puck with a mass of 3.50×10−2 kg, sliding with a speed of 0.200 m/s on a frictionless, horizontal air table, makes a perfectly elastic, head-on collision with a red puck with mass m(red) , initially at rest. After the collision, the velocity of the blue puck is 4.00×10−2 m/s in the same direction as its initial velocity.

a. Find the velocity v(red) of the red puck after the collision.



b. Find the mass m(red) of the red puck.

Answer 1

In thoughtdream's 3rd paragraph he assumes the masses are equal. That's the only way he could find Vr so easily.









I spent a while on this last night and here are the results:



The velocity of the center of mass of Mb & Mr combined is the same before and after the collision.

Vc = (MbVb + MrVr)/(Mb + Mr) = .007/(.035+Mr) ...........(1



The change in momentum of blue = the negative of the change in red:

∆Pb = -∆Pr → Pr = -.035(.04-.20) = .0056 = MrVr............(2



The velocity at which Mr recedes from the center of mass is

Ur = Vb/(1+Mr/Mb). Therefore, using (1,



Vr = Vc + Ur = .007/(.035+Mr) + .20/(1+Mr/.035) .............(3



By putting (3 into (2 gives an eq in Mr only which can be solved. I got Mr = .02333 kg.

Then use (2 again to find that Vr = .24 m/s



To check this, the total initial momentum is .035*.20 = .007

Final momentum is .035*.04 + .02333*.24 = .007 kg.m/s

Answer 2

Two key things in this problem, frictionless surface & perfectly elastic collision.



Because the collision is perfectly elastic one can assume the velocity remains constant. The original pre-collision velocity should equal the total post-collision velocities of the pucks.



So the original blue puch velocity of 0.200m/s would equal the post-collision velocity of the blue puck (4.00x10-2m/s) + the post-collision velocity of the red puck. Subtract to to find the velocity of the red puck.



To find the mass, use 'conservation of momentum'. The momentum pre-collision remains the same, post-collision.



Momentum equals mass * velocity.



Momentum pre-collision = TOTAL momentum post-collision...



pre-collision blue puck momentum = post-collision blue puck momentum + post-collision red puck momentum.



Momentum pre-collision is the mass of the blue puck (3.50x10-2kg) * the velocity of the blue puck (0.200m/s).



Post collision there are two pieces of momentum. The post-collision blue puck has a new momentum of its mass (3.50x10-2) * its NEW velocity (4.00x10-2m/s).



The post-collision red puck has a moment of its mass (unknown) * its NEW velocity (figured out in the first part).



So now you have all the pieces defined except the red pucks mass.



pre-collision=post-collision

mv(blue) = mv (blue) + mv (red)

(3.50x10-2)(0.220)=(3.50x10-2)(4.00x10-2) + m(red)*velocity(red, from part a).



A little algebra and your all done.

Answer 3

The word "completely elastic collision" is the main it shows that the products bounce off of the different and commute in opposite guidelines. So the 440g p.c.. would be travelling west after the collision and the 870g p.c.. would be travelling east. It additionally shows that ability is conserved so which you will equate kinetic ability earlier and after the collision.