Question #1e5b4

1 Answer
Nathan L.
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).

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