Question

(b) Fig. 3.2 shows a model train, travelling at speed v, approaching a buffer.
model train
buffer
spring
Fig. 3.2
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 0.48J.
Calculate the initial speed v of the train.
VE
[4]
[Total: 6]
Page
2
11

Answer 1

Answer:

K.E. = 1/2 x m x v^2

0.48J = 1/2 x 2.5 x v^2

0.48J = 1.25v^2

v^2 = 0.48 / 1.25

v^2 = 0.384 m/s

Explanation:

Answer 2

Given:

The mass of the train = 2.5 kg

The energy stored in the spring when the train stopped = 0.48 J

To find:

Find the initial speed v of the train.

Solution:

It is given that the train traveling with a velocity 'v' is stopped by a buffer spring. After the train is stopped completely the energy stored in the spring is 0.48 J.

So, we can see here that the kinetic energy of the train is converted into the potential energy stored in the buffer spring. Therefore

K.E of spring = 0.48 J

$\frac{1}{2}mv^2$ = 0.48

v² = 2 x 0.48 / 2.5

v² = 0.384

v = 0.619 m/s

Therefore the initial velocity of the train will be 0.619 m/s.