Question

Two blocks are free to slide along the frictionless wooden track shown below. The block of mass m₁ = 5.07 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 = 10.3 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m₁ rises after the elastic collision.

Answer 1

ELASTIC COLLISION

m1 = 5.07 kg                                     m2 = 10.3 kg

speeds before collision

v1i = sqrt(2*g*h1) = sqrt(2*9.81*5) =9.9 m/s         v2i = -v1

speeds after collision

v1f =                                                 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)

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

------------------------------

v1f = ((5.07 - 10.3)*9.9 + (2*10.3*0))/(5.07+10.3)

v1f = -3.37 m/s

maximum height to which m1 rises H = v1f^2/92*g)

H = 3.37^2/(2*9.81)

H = 0.58 m