More about collisions
1. Maximum angle of deflection
In a collision perfectly elastica, a moving particle of mass M collides
with a stationary particle of mass m < M.
We are interested to determine the maximum possible angle θ
through which the incident particle can be deflected.
Let's call the velocity of the moving particle v before the collision and
w after the collision; call the velocity of the originally stationary particle
u after the collision.
We seek the maximum angle of deflection is the angle between v and w.
According to:
The conservation of momentum:
Mv = Mw + mu (1)
The conservation of kinetic energy:
(1/2)Mv2 = (1/2)Mw2 + (1/2)mu2 (2)
The equation (1) gives :
u = M(v - w)/m
The equation (2) becomes then:
Mv2 = Mw2 + m[M(v - w)/m]2
mM(v2 - w2) = M2(v2 - 2 v.w + w2)
2 v.w = w2 + v2 - (m/M)(v2 - w2)
2 M v.w = (M + m)w2 + (M - m)v2
cos θ = [(M + m)w2 + (M - m)v2] /(2Mvw) =
(M + m)w/2Mv + (M - m)v/2Mw (3)
Differentiating with respect to w, and solve
for zero gives:
0 = (M + m)/2Mv - (M - m)v/2Mw2
Therefore:
w = v [(M - m)/(M + m)]1/2
w = v [(M - m)/(M + m)]1/2
Substituting this to the equation of cosθ yields:
cos θ = (M + m)w/2Mv + (M - m)v/2Mw =
(1/2M)[(M + m)w/v + (M - m)v/w] =
(1/2M)[(M + m)[(M - m)/(M + m)]1/2 + (M - m)[(M + m)/(M - m)]1/2] =
(1/2M)[(M + m)1/2(M - m) + (M - m)1/2(M + m)] =
[(1 + m/M)(1 - m/M)] 1/2
Therefore:
cos θ = [(1 + m/M)(1 - m/M)] 1/2
cos θ = [(1 + m/M)(1 - m/M)] 1/2
Hence:
sin2θ = 1 - cos2θ =
1 - [(1 + m/M)(1 - m/M)] = (m/M)2
Therefore:
sinθ = (m/M)
The maximum angle of deflection
θ is:
θ = sin-1(m/M)
Maximum angle of deflection:
θmax = sin-1(m/M)
2. Head-on collision
Let's rewrite the formula (3):
cos θ = (M + m)w/2Mv + (M - m)v/2Mw
Calling (M + m)/2M = S, and (M - m)/2M = D, we have:
cos θ = S w/v + Dv/w
vw cos θ = Sw2 + Dv2
We obtain the following quadratic equation:
Sw2 - (v cosθ)w + Dv2 = 0
Sw2 - (v cosθ)w + Dv2 = 0
The solutions are:
w = {(v cosθ) +/- [Δ]1/2 }/2S
Δ = (v cosθ)2 - 4 SDv2
The ddifferentiation of w with respect to θ is:
dw/dθ = {- (v sin θ) +/- d[Δ]1/2/dθ}/2S
d[Δ]1/2/dθ = (1/2) dΔ/dθ/[Δ]1/2
dw/dθ = {- (v sin θ) +/- ((1/2) dΔ/dθ/[Δ]1/2)}/2S
dΔ/dθ = - 2v2cosθ sinθ
dw/dθ = {- (v sin θ) +/- ((1/2) (- 2v2cosθ sinθ)/[Δ]1/2)}/2S
= {- (v sin θ) -/+ ((v2cosθ sinθ)/[Δ]1/2)}/2S
Equating to zero gives:
(v sin θ) = -/+ ((v2cosθ sinθ)/[Δ]1/2)
We have then:
sin θ = 0, that is θ = 0, and
Δ = v2 cos2θ
1 = -/+ ((v cosθ/[Δ]1/2)
Δ = v2cos2θ. That is:
(v cosθ)2 - 4 SDv2 = v2cos2θ.
Or:
4 SDv2 = 0, that is D = 0 : M = m
Conclusion:
Whem θ = 0 , the speed w, then the kinetic energy
(1/2)Mw2, of the mass M after collision
is maximum, that is the collision is head-on.
3. Maximum energy transfer
From the conservation of momentum:
Mv = Mw + mu , and
The conservation of kinetic energy:
(1/2)Mv2 = (1/2)Mw2 + (1/2)mu2
we write:
u = (M/m)(v - w)
v2 = w2 + (M/m) (v2 + w2 -2vw)
Let S = M + m, and D = M - m, we have then:
sw2 - 2(Mv)w + Dv2 = 0
Δ' = M2v2 - SDv2 = m2v2
The solution are:
w1 = (Mv + mv)/S = v
w2 = (Mv + mv)/S = v (M - m)/(M + m) (4)
The two solutions means that if M = m, the
mass M stops (w2 = 0) and the mass m takes its
velocity ( w1 = v). we have here an excahnge
of velocity (then kinetic energy).
The maximum energy transterred to the mass m is then:
Tmax = (1/2)Mv2 - (1/2)Mw2 = (1/2)mu2
According to the relationship (4), we have:
Tmax = (1/2)Mv2 - (1/2)Mw2 =
(1/2)Mv2 - (1/2)M(v (M - m)/(M + m))2 =
(1/2)Mv2 [1 - ((M - m)/(M + m))2 ] =
(1/2)Mv2 [(4mM)/(M + m))2 ]
Writing as To = (1/2)Mv2 as the initial energy of the
projectile, then:
Tmax = (1/2)Mv2 [(4mM)/(M + m))2] =
To [(4mM)/(M + m))2]
Initial energy of the projectile M:
To = (1/2)Mv2
Max energy transferred to the traget m:
Tmax = 4mMTo/(M + m)2
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