Question 1: From the reference frame of the earth, two balls (one of mass m and

Question

Question 1: From the reference frame of the earth, two balls (one of mass m and the other of mass M) are moving horizontally towards one another. The small ball, m, is moving to the right (+i) with a speed of v, and the large ball is traveling to the left (-i) with a speed of u. The bals elastically collide and stay on the r axis after the collisions. a) Calculate the speed and direction the balls travel off with after the collision; assuM (1 - Zu). Momentum and Kinetic energy would be great starting locations for this problem.) (Hint b) Get numerical values for the final speeds of the balls given: M- 3kg, m 1kg, 3, and5 c) Use a Galilean coordinate transformation to go into the large ball's point of view before the collision and get the velocity of the sa ball and large ball before and after the collision. You can just use your numerical results from part (b), the algebra gets rather hairy otherwise. Remember your directions! d) Using the numbers you got from part (c), check to see if the momentum and kinetic energy is conserved during the collision in the large ball's reference frame. Will the equations you used for momentum and kinetic energy worlk for special relativistic collisions? Explain in 1 or 2 sentences.

Explanation / Answer

given from the reference frame of earth

two balls, m and M

velocity of m = vi

velocity of M = -ui

elastic 1 d collision

a. after collision let speed of the balls be v' and u' respectively

then from conservation of moemntum

mv - Mu = mv' + Mu'

v' = (mv - Mu - Mu')/m

also

from conservation of energy

mv^2 + Mu^2 = mv'^2 + Mu'^2

mv^2 + Mu^2 = [m^2v^2 + M^2*u^2 + M^2u'^2 - 2Mmvu + 2M^2*uu' - 2Mmvu' + Mmu'^2]/m

Mu^2 = [ M^2*u^2/m + M^2u'^2/m - 2Mvu + 2M^2*uu'/m - 2Mvu' + Mu'^2]

u'^2(M + M^2/m) + u'(-2Mv + 2M^2*u/m) + (M^2u^2/m - 2Mvu - Mu^2) = 0

u' = [(2Mv - 2M^2*u/m) +- sqrt((2Mv - 2M^2*u/m)^2 - 4(M + M^2/m)(M^2u^2/m - 2Mvu - Mu^2))]/2(M + M^2/m)

and

v' = (mv - Mu - Mu')/m

b. M = 3 kg

m = 1 kg

v = 3 m/s

u = 5 m/s

hence

u' = [(2Mv - 2M^2*u/m) +- sqrt((2Mv - 2M^2*u/m)^2 - 4(M + M^2/m)(M^2u^2/m - 2Mvu - Mu^2))]/2(M + M^2/m)

u' = [(2*3*3 - 2*3^2*5) +- sqrt((2*3*3 - 2*3^2*5)^2 - 4(3 + 3^2)(3^2*5^2 - 2*3*3*5 - 3*5^2))]/2(3 + 3^2)

u' = (-72 +- 48)/24 = -1, -5 m/s

v' = (3 - 3*5 - 3*u') = -9, 3 m/s

intitial momentum = 1*3 - 3*5 = 3 - 15 = -12 kg m/s

final momentum

case 1 : 1*(-9) + 3*(-1) = -12 kg m/s

case 2 : 1*(3) + 3*(-5) = 3 - 15 = -12 kgm/s

initial energy = 0.5*(1*9 + 3*25) = 42 J

final energy

case 1 : 0.5(1*81 + 3*1) = 42 J

case 2: 0.5(1*9 + 3*25) = 42 J

now this means

there are two possible answers

u', v' = -1. -9 m/s

u',v'= 3, -5 m/s

c. in large balls point of view

before collision:

speed of large ball = 0 m/s

speed of small balll = -3 m/s ( towards )

after collision

speed of large ball = 0 m/s

speed of small ball =

case 1 = 8 m/s ( away)

case 2 = 8 m/s away

d. we have proved energy conservation in part 2 and it will always hold in all fgrames of refgerences

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