Question #76149
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
Expert's answer
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
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
