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 5.08 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 10.7 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which mn rises after the elastic collision. If m1 returns to its original height it will have the same gravitational potential energy. If block m2 is also moving at this time, what does that say about the total mechanical energy in the system? m m1

Explanation / Answer

ELASTIC COLLISION

m1 = 5.08 kg

m2 = 10.7 kg

velocity before collision

velocity of m1 = v1i = sqrt(2*g*h) = sqrt(2*10*5) = 10 m/s

velocity of m2 = v2i = 0

velocity after collision

velocity of m1 = v1f

velocity of m2 = v2f

initial momentum before collision

Pi = m1*v1i + m2*v2i

after collision final momentum

Pf = m1*v1f + m2*v2f

from momentum conservation

total momentum is conserved

Pf = Pi

m1*v1i + m2*v2i = m1*v1f + m2*v2f .....(1)

from energy conservation

total kinetic energy before collision = total kinetic energy after collision

KEi = 0.5*m1*v1i^2 + 0.5*m2*v2i^2

KEf =   0.5*m1*v1f^2 + 0.5*m2*v2f^2

KEi = KEf

0.5*m1*v1i^2 + 0.5*m2*v2i^2 = 0.5*m1*v1f^2 + 0.5*m2*v2f^2 .....(2)

solving 1&2

we get

v1f = ((m1-m2)*v1i + (2*m2*v2i))/(m1+m2)

v1f = ((5.08-10.7)*10)/(5.08+10.7)

v1f = -3.56 m/s

maimum height hmax = v1f^2/(2*g) = 3.56^2/(2*10) = 0.634 m

==================================

v2f = ((m2-m1)*v2i + (2*m1*v1i))/(m1+m2)

v2f = ( (10.7-5.08)*0 + (2*5.08*10))/(5.08+10.7)

v2f = 6.44 m/s

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