Question

Question #1e5b4

Answer 1

`v_1 = 4.83` (no discernible units given, although it probably is in `"m/s"`)

Explanation:

The formula for kinetic energy is

`KE = 1/2mv^2`

We also know that the kinetic energy doubled:

`KE_2 = 2KE_1`

and that the final speed `v` of the particle is two units more than than the original speed:

`v_2 = v_1 + 2`

The equations we can write for each situation are

`KE_1 = 1/2m(v_1)^2`

`KE_2 = 1/2m(v_2)^2`

If we rearrange the first equation, we find that

`2KE_1 = m(v_1)^2`

We found earlier that `2KE_1 = KE_2`, so this is thus equal to the second equation:

`color(red)(m(v_1)^2 = 1/2m(v_2)^2`

Now we're solving for the initial speed, `v_1`, and we also know that `v_2 = v_1 + 2`:

`m(v_1)^2 = 1/2m(v_1+2)^2`

`(v_1)^2 = 1/2(v_1+2)^2`

`(v_1)^2 = 1/2((v_1)^2 + 4v_1 + 4)`

`2(v_1)^2 = (v_1)^2 + 4v_1 + 4`

`ul((v_1)^2 - 4v_1 - 4 = 0`

Using the quadratic equation yields:

`color(blue)(ulbar(|stackrel(" ")(" "v_1 = 4.83color(white)(l)"LT"^-1" ")|)`

(taking the positive solution).

(The symbol `"LT"^-1` is the dimensional form of the units for velocity).