Q 1: Two masses are on separate sides of a U-shaped track. Mass #1 is 2.2 kg and

Question

Q 1: Two masses are on separate sides of a U-shaped track. Mass #1 is 2.2 kg and is located 4.0 m above the bottom of the track. Mass #2 is 7.0 kg and is located 2.0 m above the bottom of the track on the other side of the U. Using only work, energy, and momentum, determine velocity of each mass after an ELASTIC collision. Q2: Two masses are on separate sides of a U-shaped track. Mass #1 is 2.2 kg and is located 4.0 m above the bottom of the track. Mass #2 is 7.0 kg and is located 2.0 m above the bottom of the track on the other side of the U. Using only work, energy, and momentum, determine velocity of each mass after an INELASTIC collision. Q3: For each question, what happens if you... a) Increase the starting height of Mass #1 b) Decrease the starting height of Mass #2 c) Increase the mass of Mass #1 d) Decrease the mass of Mass #2

Explanation / Answer

Q1)m1 = 2.2 kg ; h1 = 4 m ;

m2 = 7 kg ; h2 = 2 m

from conservation of energy,

Ke = PE

1/2 m v^2 = m g h

v = sqrt (2 g h)

v1 = sqrt (2 g h1) = sqrt (2 x 9.8 x 4) = 8.85 m/s

v2 = sqrt (2 g h2) = sqrt (2 x 9.8 x 2) = 6.26 m/s

from conservation of momentum

m1v1 + m2v2 = m1 v1' + m2 v2'

from conservation of KE

1/2 m1 v1^2 + 1/2 m2 v2^2 = 1/2 m1 v1'^2 + 1/2 m2 v2'^2

solving the above two simultaneously we get

v1' = (m1 -m2)v1/(m1 + m2) + 2m2v2/(m1 + m2)

v2' = 2m1v1/(m1 + m2) - (m1 - m2)v2/(m1 + m2)

putting in the values:

v1' = (2.2 - 7)8.85/(2.2 + 7) + 2 x 7 x 6.26/(2.2 + 7) = 4.91 m/s

v2' = 2 x 2.2 x 8.85/(2.2 + 7) - (2.2 - 7)6.26/(2.2 + 7) = 7.49 m/s

Hence, velocities after collision is:

v1' = 4.91 m/s ; v2' = 7.49 m/s

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