Two blocks are free to slide along the frictionless wooden track shown below. Th

Question

Two blocks are free to slide along the frictionless wooden track shown below. The block of mass m1 = 5.09 kg is released from the position shown, at height h = 5.00 m above the flat part of the track. Protruding from its front end is the north pole of a strong magnet, which repels the north pole of an identical magnet embedded in the back end of the block of mass m2 = 10.1 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.

?m

Explanation / Answer

This one has three basic steps. First, use conservation of energy to find the velocity of the first block just before the collision. Call the initial height h = 5 m, and let the zero for potential energy be the vertical position of m2:
Ki + Ui = Kf + Uf
0 + mgh =(1/2)mv1i2
v1i = (2gh)
Next, we handle the collision itself. It is an elastic collision, so both kinetic energy and momentum are conserved. We have already derived the equation for v1f for a 1D elastic collision, and only state the result below.
v1f = (m1-m2/m1+m2)v1i =[(m1-m2/m1+m2)](2gh)

The last step is to use conservation of energy again to nd the new height, which we'll call h'.

(1/2)m1v1f2 =m1gh'

h' =v1f2/2g =(m1-m2/m1+m2)2(2gh)*2g

h'=h[(m1-m2/m1+m2)2]

h'=5[5.09-10.1/5.09+10.1]2

h'=0.544m

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