2 trolleys A and Bof masses 0.6 and 0.40,respectively are held together on a horizontal runway againsta spring which is in a state of compression. when the spring is releases, the trolleys seperate freely and A moves to the right with initial velocity of 5.0m/s
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Expert's answer
2018-04-18T08:13:07-0400
From the conservation of momentum, Sum of momentum before collision = Sum of momentum after collision. Before collision, the sum of momentum is zero since the trolleys are stationary. After collision, 0.6 (5) + 0.4 (v) = 0 Velocity v = -7.5 m/s. As the trolleys move, they have kinetic energy. This kinetic energy was initially stored in the spring. Energy stored = Sum of kinetic energies of trolleys E=1/2 0.6 (5)^2+ 1/2 0.4 (-7.5)^2=19 J
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