Question

The proton with initial v1 = 3.00 x 10^5 m/s enters a region 1.00 cm long where it is accelerated. Its final velocity is v = 6.40 x 10^6 m/s . What was its constant acceleration. Answer in 10^15 m/s^2

Answer 1

The constant acceleration,
`a_(c) = 2.044* 10^15 m/s^2`

Solution:

As per the question:

`v_(1) = 3 x 10^5 m/s`

`v_(2) = 6.40 x 10^6 m/s`

length of the re region, l = 1 cm = 0.01 m

Now,

Using the third equation of motion:

`v_(2)^2 = v_(1)^2 + 2a_(c)l`

`a_(c) = (v_(2)^2 - v_(1)^2)/(2l)`

`a_(c) = ((6.40 x 10^6)^2 - (3 x 10^5)^2)/(2* 0.01)`

`a_(c) = 2.044* 10^15 m/s^2`