|Share this thread:|
Nov2-05, 11:31 PM
(3) two blocks are free to slide along the frictionless wooden track. a block of mass m1=5.00kg is released from A. Protruding from its front end is the north pole of a strong magnet, repelling the north pole of an identical magnet embedded in the back end of the block of mass m2=10.0kg, initially at rest. the two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.
Do you use conservation of kinetic energy and momentum to figure out the final velocity of the block M1 after the collision. Then let all that kinetic energy be converted to gravitational potential energy to determine how high it rises???
So GPE = KE , mgh= 1/2mv^2
if (5)(9.81)(5) = 1/2 (10)v^2
u find initial velocity?...
do u sub that into
vfinal = (m1-m2)/(m1 + m2) *vinitial?
to find vfinal...?
and then sub that back into...
mgh = 1/2 mv^2
to solve for h=1/2mv^2/mg?
OR iS THis completely wrong? somebody let me know if im on the right track.. or where i veered off the track...
Nov3-05, 07:30 AM
I think you did something wrong there. In your conservation of momentum part, you seem to have ignored completely the velocity of the 2nd mass that was stationary (m2) after the collision. So I'm not sure how you got vfinal.
Let's do this systematically.
1. m1 sliding down. This part is OK. So using the conservation of energy, you have v_i = 2gh.
2. So now m1 will hit m2 with an initial velocity of v_i. After impact, both m1 and m2 will move. So use conservation of momentum, i.e.
m1*v_i = m1*v_f + m2*v2
Notice you have 1 equation but 2 unknowns, v_f and v2.
3. But notice that this is an elastic collision. There's only magnetic interactions and no energy in the mechanics is coverted into other means, so you have another equation to describe the collision, the conservation of KE, i.e.
1/2 m1*v_i^2 = ...... etc. (you should know how to write this).
4. Using (2) and (3), you now have 2 equations and 2 unknowns. You should be able to solve for v_f.
5. Using v_f, you know the KE of m1 as it goes back climbing that slope. Again use conservation of energy to find out the height it will get.
And interesting side question would be to find the value of m2 in which m1 will no longer bounce back, i.e. after collision, m1 and m2 will move in the same direction.
Nov4-05, 01:26 AM
What i seem to get was the same thing as you...but i got a little confused with your writing so hopefully we are saying the same thing:
v1f (final velocity of mass 1) = [(m1-m2)/(m1+m2)]*v1i
then to find the max height,
m1ghmax = (1/2)m1v1f^2
so, you can see that m1 cancels. therefore, you're left with
hmax = v1f^2/(2g)
hopefully that made some sense.
|Register to reply|
|Collision & Momentum||Introductory Physics Homework||10|
|Momentum and Collision||Introductory Physics Homework||2|
|Momentum and Collision||Introductory Physics Homework||6|
|Momentum after collision||Introductory Physics Homework||2|
|Mastering Physics: Collision on an incline plane and perfectly inelastic collision||Introductory Physics Homework||2|