Three balls each have a mass m if a has a speed v just before a direct collision with B determine the speed of C after collision the coefficient of restitution between each pair of balls is e. neglect the size of each ball

(vb)_3 = [v(1 - e)²]/4

(vc)_2 = [v(1 + e)²]/4

Explanation:

From conservation of momentum, when ball A strikes ball B, we have;

ma(va)_1 + mb(vb)_1 = ma(va)_2 + mb(vb)_2

Ball B is at rest before being hit. so (vb)_1 = 0 m/s

Thus, after being hit, we now have;

mv = m(va)_2 + m(vb)_2 - - - (eq1)

Coefficient of restitution is given as;

e =[((vb)_2) - ((va)_2)]/[((va)_1) - ((vb)_1)]

(vb)_1 = 0 m/s

Thus,we now have;

ve =[((vb)_2) - ((va)_2)] - - - (eq2)

If we divide each term in eq 1 by m, we have;

v = (va)_2 + (vb)_2 - - - - eq3

Now, let's put add eq 3 to eq 2 to obtain ;

v + ve = 2(vb)_2

v(1 + e) = 2(vb)_2

(vb)_2 = [v(1 + e)]/2 - - - - - eq(4)

Now let's subtract eq 2 from eq 3 to obtain;

v - ve = (va)_2 + (va)_2

v(1 - e) = 2(va)_2

(va)_2 = [v(1 - e)]/2

Now let's repeat the same process for conservation of momentum when ball B strikes ball C;

mb(vb)_2 + mc(vc)_1 = mb(vb)_3 + mc(vc)_2

Ball C is at rest before being hit. so (vc)_1 = 0 m/s

Thus, after being hit, we now have;

m(vb)_2 = m(vb)_3 + m(vc)_2

From earlier, (vb)_2 = [v(1 + e)]/2

Thus, we now have;

m[v(1 + e)]/2] = m(vb)_3 + m(vc)_2

Divide through by m;

[v(1 + e)]/2] = (vb)_3 + (vc)_2 - - - eq5

Coefficient of restitution is given as;

e =[(vc)_2) - ((vb)_3)]/[((vb)_2) - ((vc)_1)]

(vc)_1 = 0 m/s, so

e =[(vc)_2) - ((vb)_3)]/[((vb)_2)]

From earlier, (vb)_2 = [v(1 + e)]/2

Thus,

e =[(vc)_2) - ((vb)_3)]/ [v(1 + e)]/2]

e[v(1 + e)/2] = [(vc)_2) - ((vb)_3)]

- - - eq6

Let's add eq 5 to eq6;

[v(1 + e)/2](1 + e) = 2(vc)_2

2(vc)_2 = [v(1 + e)²]/2

(vc)_2 = [v(1 + e)²]/4

Also,let's subtract eq 6 from eq5 to obtain;

[v(1 + e)]/2](1-e)= 2(vb)_3

2(vb)_3 = [v(1 - e)²]/2

(vb)_3 = [v(1 - e)²]/4

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