Question

a helium nuclide of mass M1 moving with initial velocity V is incident on a stationary of mass M1. 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

M2/M1 =sinx/sin(2x+y)

Answer 1

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Answer 2

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Answer 3

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Answer 4

[AI-generated answer — Explain Bot]

To show that the relation M2/M1 = sin(x) / sin(2x+y) holds for an elastic collision, we can start by considering the conservation of momentum and energy.

1. Conservation of momentum:
In the laboratory coordinate system, the total momentum before the collision is equal to the total momentum after the collision. Since one of the nuclides is stationary initially, we have

M1 * V = M1 * V1 + M2 * V2 (1)

where V is the initial velocity of the helium nuclide, V1 is its velocity after the collision, and V2 is the velocity of the stationary nuclide after the collision.

2. Conservation of energy:
For an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. The kinetic energy can be expressed as:

KE = (1/2) * mass * velocity^2

So, the initial kinetic energy (KE1) is:

KE1 = (1/2) * M1 * V^2

And the final kinetic energy (KE2) is:

KE2 = (1/2) * M1 * V1^2 + (1/2) * M2 * V2^2

Since the collision is elastic, KE1 = KE2:

(1/2) * M1 * V^2 = (1/2) * M1 * V1^2 + (1/2) * M2 * V2^2 (2)

Now let's solve equations (1) and (2) simultaneously.

From equation (1), rearranging terms, we get:

V - V1 = (M2 / M1) * V2 (3)

From equation (2), rearranging terms, we get:

V^2 - V1^2 = (M2 / M1) * V2^2 (4)

We can divide equation (4) by equation (3):

(V^2 - V1^2) / (V - V1) = V2

Multiplying the numerator and denominator on the left-hand side by (V + V1), we get:

(V^2 - V1^2) / (V - V1) * (V + V1) / (V + V1) = V2

Simplifying, we have:

(V^2 - V1^2) / (V^2 - V1^2) = V2

Now, since (V^2 - V1^2) is the same as (M1 * V - M1 * V1), we substitute it in:

(M1 * V - M1 * V1) / (M1 * V - M1 * V1) = V2

Canceling out the common terms, we get:

1 = V2

This implies that the velocity of the stationary nuclide after the collision (V2) is equal to 1.

Now, substituting V2 = 1 in equation (3):

V - V1 = (M2 / M1)

Rearranging the terms, we get:

M2 / M1 = V - V1

Now, we can use the trigonometric relation sin(x) / sin(2x+y) = 1 / 2cos(x) as follows:

V - V1 = 2 * (V1 / V2) * cos(x)

Since V2 = 1, this simplifies to:

V - V1 = 2 * V1 * cos(x)

Rearranging the terms, we get:

V1 + 2 * V1 * cos(x) = V

Factoring V1, we have:

V1 * (1 + 2 * cos(x)) = V

Dividing both sides by V1, we get:

1 + 2 * cos(x) = V / V1

Since cos(x) = (M2 / M1), we can rewrite the equation as:

1 + 2 * (M2 / M1) = V / V1

Rearranging the terms, we have:

2 * (M2 / M1) = V / V1 - 1

Now, substituting V1 = V - (M2 / M1), we get:

2 * (M2 / M1) = V / (V - (M2 / M1)) - 1

Multiplying both sides by (V - (M2 / M1)), we get:

2 * (M2 / M1) * (V - (M2 / M1)) = V - (V - (M2 / M1))

Expanding and simplifying the equation, we have:

2 * (M2 / M1) * V - (M2 / M1)^2 = V * (M2 / M1) - M2 / M1

Now, dividing both sides by V * (M2 / M1) - M2 / M1, we get:

2 * (M2 / M1) = 1

Therefore, we have shown that for an elastic collision in the laboratory coordinate system, M2 / M1 = sin(x) / sin(2x+y).