Consider a particle of mass m₁, moving with
a constant velocity u₁, which collides elastically with a stationary particle
of mass m₂ in the laboratory frame of reference.
After the collision, the particle of mass
m₁ moves with velocity v₁, making an angle θ₁ with its initial direction of
motion, while the particle of mass m₂ moves with velocity v₂, making an angle
θ₂ with the initial direction of motion of the first particle.
Let the initial direction of motion of m₁
be along the X-axis, and let the velocities v₁ and v₂ lie in the X–Y plane, as
shown in Fig. 1.
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The velocity of the centre of mass in the
laboratory frame is
VCM = m₁u₁ / (m₁
+ m₂) (1)
The initial velocity of particle m₁ in the
C.M. frame is
u₁′ = u₁ − V_C (2)
Using Eq. (1),
u₁′ = m₂u₁ / (m₁ + m₂) (3)
The initial velocity of particle m₂ in the
C.M. frame is
u₂′ = − VCM =
−m₁u₁ / (m₁ + m₂) (4)
After the collision, let the velocities of
the two particles in the C.M. frame be v₁′ and v₂′, respectively.
Since the centre of mass remains at rest in
the C.M. frame, the total momentum before and after collision is zero.
Therefore,
m₁u₁′ + m₂u₂′ = 0 (5)
u₂′ = −(m₁/m₂)u₁′ (6)
Similarly, after collision,
m₁v₁′ + m₂v₂′ = 0 (7)
v₂′ = −(m₁/m₂)v₁′ (8)
Equation (8) shows that after the
collision, the two particles move in opposite directions in the C.M. frame,
as shown in Fig. 2.
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Fig. 2: Elastic collision in the
centre-of-mass frame |
Since the collision is elastic,
conservation of kinetic energy in the C.M. frame gives
½m₁u₁′² + ½m₂u₂′² = ½m₁v₁′²
+ ½m₂v₂′² (9)
Substituting Eqs. (6) and (8) into Eq. (9),
½m₁u₁′² + ½m₂(m₁u₁′/m₂)² =
½m₁v₁′² + ½m₂(m₁v₁′/m₂)² (10)
m₁u₁′²(1 + m₁/m₂) =
m₁v₁′²(1 + m₁/m₂) (11)
v₁′² = u₁′² ⟹ v₁′
= u₁′ (12)
v₂′ = u₂′ (13)
Thus, the magnitudes of the velocities of
both particles in the C.M. frame remain unchanged after an elastic collision.
However, their directions may change.
Value of the Scattering Angle
In the C.M. frame, the scattering angle can
have any value. Therefore, there is no restriction on the scattering angle in
this frame. However, in the laboratory frame, the scattering angle θ₁ is
restricted by the mass ratio of the two particles.
From Fig. 9.1,
tan θ₁ = (v₁ sin θ)/(v₁ cos
θ) (14)
The transformation from the C.M. frame to
the laboratory frame gives
v₁ sin θ₁ = v₁′ sin θ (15)
v₁ cos θ₁ = v₁′ cos θ + VCM (16)
tan θ₁ = v₁′ sin θ / (v₁′
cos θ + VCM ) (17)
From Eq. (4),
VCM = m₁u₁/(m₁ +
m₂) (18)
and from Eq. (6),
u₁′ = m₂u₁/(m₁ + m₂) (19)
Hence,
VCM = (m₁/m₂)u₁′ (20)
Since v₁′ = u₁′,
VCM = (m₁/m₂)v₁′ (21)
Substituting Eq. (24) into Eq. (20),
tan θ₁ = sin θ / [cos θ +
(m₁/m₂)] (22)
This equation determines the possible range
of the scattering angle θ₁ in the laboratory frame.
(a) When m₁ > m₂
If m₁/m₂ > 1, the denominator of Eq. (22)
cannot become zero because cos θ + m₁/m₂ > 0. Therefore,
0 ≤ θ₁ < π/2 (23)
Thus, when a heavy particle collides
elastically with a lighter stationary particle, the heavy particle cannot be
scattered backward. This case is shown in Fig. 9.3(a).
(b) When m₁ = m₂
If m₁/m₂ = 1, the denominator of Eq. (22)
becomes zero when cos θ = −1, i.e., θ = π. In this limiting case,
θ₁ = π/2 (24)
Thus, when two identical particles collide
elastically and one particle is initially stationary, the scattered particle
can move through an angle up to 90° with respect to its initial direction.
Hence,
0 ≤ θ₁ ≤ π/2 (25)
This case is shown in Fig. 9.3(b).
(c) When m₁ < m₂
If m₁/m₂ < 1, the denominator of Eq. (22)
can become negative. Consequently, tan θ₁ can be negative and the scattering
angle can exceed 90°.
Thus, when a light particle collides
elastically with a heavier stationary particle, the lighter particle may be
scattered backward.
0 ≤ θ₁ ≤ π (26)