in an atomic physics experiment, two atoms collide with each other in such a way
ID: 2175665 • Letter: I
Question
in an atomic physics experiment, two atoms collide with each other in such a way that the initial total momentum of the atoms is zero. The atoms are chosen so that M2=2M1.In a collision, the atom M1 has a final velocity V1f=-1/3(V1i). What is the realation between the final kinetic energy and the initial kinetic energy of the system (the two atoms)?
a) KEf=KEi
b) KEf=1/2KEi
c) KEf=1/2KEi
d) KEf=1/4KEi
e) KEf=1/6KEi
f) KEf=1/9KEi
g) KEf=1/sqrt(2) KEi
h) KEf= 1/sqrt(3) KEi
i) KEf=1/sqrt(6) KEi
j) KEi=0
Explanation / Answer
b4 colision by conservation of momentum
M1*V1i + M2*V2i = 0
=> V1i =-M2*V2i /M1= -2*V2i
after colision by conservation of momentum
M1*V1f + M2*V2f = 0
=> V2f = -M1*V1f /M2 = -M1*(-1/3(V1i)) /M2 (since given ,V1f=-1/3(V1i))
=> V2f =-(1/6)*V1i =-(1/3)*V2i (since V1i =-2*V2i)
KEi = 0.5*(M1*V1i2 + M2*V2i2)
KEf = 0.5*(M1*Vf1f2 + M2*V2f2) = 0.5*(M1*(-1/3(V1i))2 + M2*(-(1/3)*V2i)2)
So KEf = (1/9)*KEi
Ans-f
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