Two billiard balls of identical mass move toward each other. Assume that the collision between them is perfectly elastic. If the
2 answers:
Answer:
As yo may know, in a perfectly elastic collision the kinetic energy remains constant; we have that:
K1 = (1/2)M*v1^2 = M*(30^2)/2
K2 = M*(-20^2)/2
Kt = k1 + k2 = M*650
Where M is the mass of the balls:
After the collision, (I denote the new quantities with an ' symbol) we must have the same energy:
K1' = (1/2)M(v1'^2 + v2'^2) = M650
v1'^2 + v2'^2 = 1300
And we also know that the total moment must be conserved:
M*30 - M*20 = M*10 = M( v1' + v2')
10 = v1' + v2'
Now we have two equations:
10 = v1' + v2'
v1'^2 + v2'^2 = 1300
We can writhe the first equation as:
10 - v1' = v2'
and replace it in the second equation:
v1'^2 + (10 - v1')^2 = 1300
and solve it for v1:
v1'^2 + 100 + v1^2 -20v1' - 1200 = 0
2*v1'^2 - 20v1' -1200 = 0
now we must solve this for v1 with the quadratic formula:
So we have two solutions:
v1' = (5 + 25)cm/s = 30cm/s
v1' = (5 - 20)cm/s = -20cm/s
and you can find that is the same case for v2' (interchanged, when v1' = 30cm/s we have that v2' = -20cm/s, and viseverce)
Now, after the impact, the first ball must change the direction of movement (It cant keep moving forward because there is the other ball in the path) so the velocities of the balls must be interchanged, this mens that v1' = -20cm/s and v2'= 30cm/s
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