v = √ { 2*(KE) ] / m } ;
Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"] ;
and solve for "v".
______________________________________________________
Explanation:
_____________________________________________________
The formula is: KE = (½) * (m) * (v²) ;
_____________________________________
"Kinetic energy" = (½) * (mass) * (velocity , "squared")
________________________________________________
Note: Velocity is similar to speed, in that velocity means "speed and direction"; however, if you "square" a negative number, you will get a "positive"; since: a "negative" multiplied by a "negative" equals a "positive".
____________________________________________
So, we have the formula:
___________________________________
KE = (½) * (m) * (v²) ; to solve for "(v)" ; velocity, which is very similar to the "speed";
___________________________________________________
we arrange the formula ;
__________________________________________________
(KE) = (½) * (m) * (v²) ; ↔ (½)*(m)* (v²) = (KE) ;
___________________________________________________
→ We have: (½)*(m)* (v²) = (KE) ; we isolate, "m" (mass) on one side of the equation:
______________________________________________________
→ We divide each side of the equation by: "[(½)* (m)]" ;
___________________________________________________
→ [ (½)*(m)*(v²) ] / [(½)* (m)] = (KE) / [(½)* (m)]<span> ;
</span>______________________________________________________
to get:
______________________________________________________
→ v² = (KE) / [(½)* (m)]
→ v² = 2 KE / m
_______________________________________________________
Take the "square root" of each side of the equation ;
_______________________________________________________
→ √ (v²) = √ { 2*(KE) ] / m }
________________________________________________________
→ v = √ { 2*(KE) ] / m } ;
Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"];
and solve for "v".
______________________________________________________
Explanation:
By the second law of Newton we get the relation
F = ma
Assume that the small-massed particle is 
and the heavier mass particle is 
.
Now, by momentum conservation and energy conservation:
Now, there are 2 solutions but, one of them is useless to this question's main point so I excluded that point. Ask me in the comments if you want the excluded solution too.
So now, we see that 
and 
. So therefore, the smaller mass recoils out.
Hope this helps you!
Bye!
