Question

A proton moving with velocity of (u)m/s and strikes a stationary nuclear of

mass (A). calculate the ratio of final to initial kinetic energy of proton.​

Answer 1

Answer:

(1-A/1+A)^2

from the elastic collision theory

Answer 2

Answer:

The ratio of final to initial kinetic energy of the proton is (1-A)²:(1+A)².

Explanation:

The final velocity of the proton is calculated as,

$v'=(\frac{m_1-m_2}{m_1+m_2})u_1+(\frac{2m_2}{m_1+m_2} )u_2$     (1)

Where,

v'=velocity of proton

m₁=mass of proton=1

m₂=mass of the stationary nucleus

u₁=initial velocity of photon

u₂=initial velocity of nucleus

From the question we have,

u₁=u

m₂=A

u₂=0           ( as the nucleus is stationary)

By substituting the values in equation (1) we get;

$v'=(\frac{1-A}{1+A})u+(\frac{2\times A}{1+A} )\times 0$

$v'=(\frac{1-A}{1+A})u$         (2)

The kinetic energy is given as,

$E=\frac{1}{2} mv^2$      (3)

E=kinetic energy of the particle

m=mass of the particle

v=velocity with which the particle is moving

The initial kinetic energy of the photon is given as,

$E=\frac{1}{2} \times 1\times u^2$      (4)

The final kinetic energy of the photon is given as,

$E'=\frac{1}{2} \times 1\times (\frac{1-A}{1+A})^2$      (5)

By taking the ratio of equation (5) upon (4) we get;

$\frac{E'}{E}= (\frac{1-A}{1+A})^2$    

Hence, the ratio of final to initial kinetic energy of the proton is (1-A)²:(1+A)².

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