Respuesta :
Answer:
Vc2= V(l+e) ^2/4
Vg2= V(l-e^2)/4
Explanation:
Conservation momentum, when ball A strikes Ball B
Where,
M= Mass
V= Velocity
Ma(VA)1+ Mg(Vg)2= Ma(Va)2+ Ma(Vg)2
MV + 0= MVg2
Coefficient of restitution =
e= (Vg)2- (Va)2/(Va)1- (Vg)1
e= (Vg)2- (Va)2/ V-0
Solving equation 1 and 2 yield
(Va)2= V(l-e) /2
(Vg)2= V(l+e)/2
Conservative momentum when ball b strikes c
Mg(Vg)2+Mc(Vc)1 = Mg(Vg)3+Mc(Vc)2
=> M[V(l+e) /2] + 0 = M(Vg)3 + M(Vc) 2
Coefficient of Restitution,
e= (Vc)2 - (Vg)2/(Vg)2- (Vc)1
=> e= (Vc)2 - (Vg)2/V(l+e) /2
Solving equation 3 and 4,
Vc2= V(l+e) ^2/4
Vg2= V(l-e^2)/4
Answer:
(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
