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astra-53 [7]
3 years ago
15

(b) The figure below shows a model train, travelling at speed , approaching a

Physics
1 answer:
Andrew [12]3 years ago
3 0

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.

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