Question #1e5b4
1 Answer
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_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
color(red)(m(v_1)^2 = 1/2m(v_2)^2
Now we're solving for the initial speed,
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