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 = 5.03 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 = 10.5 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.

Explanation / Answer

given;
m1 = 5.03kg
m2 = 10.5 kg
elastic collision ( restitutive coefficient e = 1)

principle of conservation energy

consider at free body of mass m1

Energy at A = Energy at B

m1 g h = ½ m1 v1²

v1² = 2 g h = 2(9.8)(5)

v1 = 9.8995 m/s

e = -(v1' - v2')/(v1 - v2) = 1

e = -(v1' - v2')/(9.8995 - 0) = 1

v2' = 9.8995 + v1'

equation of conserve momentum

m1 v1 + m2 v2 = m1 v1' + m2 v2'

(5.03) v1 + (10.5) (0) = (5.03) v1' + (10.5) v2'

remember v2' = 9.8995 + v1' ................so

(5.03) (9.8995) + (10.5) (0) = (5.03) v1' + (10.5) (9.8995 + v1')

v1' = -3.4868 m/s

again, conserve of energy

m g h' = ½ m (v1')²

h' = (v1')²/2g = (-3.4868)²/{2(9.8)}

h' = 0.6203 m

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