CATAGORY: Mechanics QUESTION: Two identical balls are traveling toward each other with speeds 4.0 >and 7.0 m/s, and they experience an elastic head-on collision. Obtain the >velocities (magnitude and direction) of each ball after the collision.

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Thgis is a beautiful problem involving multiple solutions that require physical analysis AFTER the math part of the problem is solved. Since the collision is elastic energy is conserved. Also momentum is ALWAYS conserved so we have. (1) KE(initial) = KE(Final) p(initial= = p(final) Where KE means kinetic energy and p means momentum. Now in general for a single particle moving with velocity v with mass M we have (2) KE = (1/2) M v^2 p = Mv Now the velocity of the first ball we will take to be moving in the positive direction so its velocity is 4.0 m/s and the second is -7.0 m/s. The second is negative because it heads towards the first ball with the opposite direction of motion of the first ball. Thus (3) KE(initial) = (1/2)M(4.0 m/s)^2 + (1/2)M(-7.0 m/s)^2 = (1/2)M( 16(m/s)^2 +49(m/s)^2) = (1/2)M 65(m/s)^2 p(initial) = M(4.0 m/s) + M(-7.0 m/s) = M(-3.0m/s) Lets call the final velocity of the first ball v1 and the second ball v2 then we have from (1) and (3) (4) (1/2)M 65(m/s)^2 = (1/2)M(v1)^2 + (1/2)M(v2)^2 M(-3.0m/s) = M v1 + M v2 Dividing through the first by (1/2)M and the second by M gives us (5) 65(m/s)^2 = (v1)^2 + (v2)^2 -3.0m/s = v1 + v2 Fronm the second equation (6) v1 = -1.0 v2 - 3.0m/s We now have an algebra problem to work. So we will drop the units for a bit. Yes this is dangerous heresy but it facilitates our work! Substitution of (6) into the first equation of (5) gives (7) 65 = (-v2 - 3)^2 + v2^2 now factoring out -1 from -v2 - 3 gives -1(v2 + 3) but (-1)^2 =1 so 65 = (v2 + 3)^2 + v2^2 expanding (v2 + 3)^2 65 = (v2)^2 + 6 v2 + 9 + (v2)^2 combining like terms 65 = 2(v2)^2 +6 v2 + 9 or 0 = 2(v2)^2 +6 v2 + 9 - 65 or 0 = 2(v2)^2 +6 v2 + - 56 apply the quadratic formula (+/- means "plus or minus" and sqrt means "square root". v2 = (-6 +/-sqrt(6^2 - 4(2)(-56))/2(2) v2 = (-6 +/-sqrt(36 + 448))/4 v2 = (-6 +/-sqrt(484))/4 v2 = (-6 +/-22)/4 = (-28/4,16/4) = (-7/4,4) Thus if v2 = -7/4 then from (6) v1 = -1 (-7/4) - 3 = 7/4 - 3 = 7/4 - 12/4 = -7/5 and if v2 = 4 then from (6) v1 = -1(4) - 3 = -7 So we have either (v1,v2) = (-7/5, -7/4) m/s or (v1,v2) = (-7, 4) m/s Now we have the pleasure of deciding which pair is physical and which is not. Note that |-7/4| is larger than |-7/5| this implies that the second ball HAS PASSED THROUGH THE FIRST BALL !!!!! Think of the first ball as initialy moving from left to right. The second moves initial from right to left as it heads to the first ball. After the collision both balls are moving to the left but the second is moving to the left faster and thus it had to pass through the first which can not happen classical so this solution is not physical! The true solution is thus ************************** * * * (v1,v2) = (-7, 4) m/s * * * ************************** the balls have completely exchanged velocities! :-)

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