Question #1e5b4

1 Answer
Aug 14, 2017

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).