Answer:
0.62 m/s
Explanation:
Given that a model train, travelling at speed , approaching a buffer model train buffer spring
The train, of mass 2.5 kg, is stopped by compressing a spring in the buffer.
After the train has stopped, the energy stored in the spring is 0.48 J.
Calculate the initial speed v of the train
Solution
The total energy stored in the spring will be equal to the kinetic energy of the train.
That is,
1/2fe = 1/2mv^2
Substitutes the spring energy, and mass into the formula
0.48 = 1/2 × 2.5 × V^2
2.5V^2 = 0.96
V^2 = 0.96 / 2.5
V^2 = 0.384
V = sqrt ( 0.384 )
V = 0.62 m/s
Therefore, the initial velocity of the train is 0.62 metres per second.
