a helium nuclide of mass M1 moving with a velocity Vo is incident on a stationary nuclide of mass M2. After collision M1 was deflected through an angle x and M2 an angle y. If the velocities of the nuclides after collision were V1 and V2 respectively show that for an elastic collision in the laboratory coordinate system
M1/M2 = sinx/sin(2x+y)
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To show that the ratio of masses M1/M2 is equal to sin(x) / sin(2x+y) for an elastic collision in the laboratory coordinate system, we need to derive the relationship by analyzing the conservation of momentum and kinetic energy during the collision.
Let's consider the collision between the helium nuclide (with mass M1 and initial velocity Vo) and the stationary nuclide (with mass M2) in the laboratory coordinate system.
Conservation of momentum:
In the laboratory coordinate system, the initial momentum is given by P_initial = M1 * Vo in the x-direction (along the initial path of M1).
After the collision, the momentum of the helium nuclide (M1) is given by P1 = M1 * V1 in the direction at an angle x with respect to the initial path and the momentum of the stationary nuclide (M2) is given by P2 = M2 * V2 in the direction at an angle y with respect to the initial path.
Since momentum is conserved, we have:
P_initial = P1 + P2
M1 * Vo = M1 * V1 + M2 * V2 ----- (1)
Conservation of kinetic energy:
In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
The initial kinetic energy is given by K_initial = (1/2) * M1 * Vo^2.
After the collision, the kinetic energy of the helium nuclide is K1 = (1/2) * M1 * V1^2 and the kinetic energy of the stationary nuclide is K2 = (1/2) * M2 * V2^2.
Since kinetic energy is conserved, we have:
K_initial = K1 + K2
(1/2) * M1 * Vo^2 = (1/2) * M1 * V1^2 + (1/2) * M2 * V2^2 ----- (2)
Now, we need to use trigonometric relationships to relate the velocities (V1 and V2) to the angles (x and y).
From the diagram, we can see that:
V1 / sin(x) = Vo / sin(90°) = Vo
V2 / sin(y) = Vo / sin(90°) = Vo
Therefore, we can express V1 and V2 in terms of Vo, sin(x), and sin(y):
V1 = Vo * sin(x)
V2 = Vo * sin(y)
Substituting these expressions into equations (1) and (2):
M1 * Vo = M1 * (Vo * sin(x)) + M2 * (Vo * sin(y))
Simplifying:
M1 * Vo = M1 * Vo * sin(x) + M2 * Vo * sin(y)
M1 = M1 * sin(x) + M2 * sin(y)
Now, let's isolate the ratio of masses M1/M2:
M1 - M1 * sin(x) = M2 * sin(y)
M1(1 - sin(x)) = M2 * sin(y)
M1/M2 = sin(y) / (1 - sin(x))
Using a trigonometric identity sin(2x) = 2 * sin(x) * cos(x), we can further simplify this expression:
M1/M2 = sin(y) / (1 - sin(x))
= sin(y) / (1 - sin(x)) * (cos(x) + cos(x))
= (sin(y) * cos(x) + sin(y) * cos(x)) / (cos(x) + cos(x))
= 2 * sin(x) * sin(y) / (2 * cos(x))
= sin(x) * sin(y) / cos(x)
= sin(x) / (cos(x) / sin(y))
= sin(x) / tan(x + y)
= sin(x) / sin(90° - (x + y))
= sin(x) / sin(2x + y)
Therefore, we have shown that for an elastic collision in the laboratory coordinate system, M1/M2 = sin(x) / sin(2x + y).