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Factors Affecting Kinetic Energy

A particle A ...

Question

A particle $A$ with a mass $m_{A}$ is moving with a velocity $v$ and hits a particle $B$ (mass $m_{B}$ ) at rest (one dimensional motion). Find the change in the de Broglie wavelength of the particle $A$ . Treat the collision as elastic.

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Solution

$Find initial momentum and kinetic energy .$

Initial momentum, $p_{i} = m_{A} v$

Initial kinetic Energy,

$K . E ._{i} = \frac{1}{2} m_{A} v^{2} + 0 = \frac{1}{2} m_{A} v^{2}$

$Find final momentum and kinetic energy .$

After collision, let the velocity of the particle $A$ be $v_{1}$ and that of particle $B$ be $v_{2}$ .

Final Momentum

$p_{f} = m_{A} v_{1} + m_{B} v_{2}$

Final Kinetic Energy

$K . E ._{f} = \frac{1}{2} m_{A} v_{1}^{2} + \frac{1}{2} m_{B} v_{2}^{2}$

$Apply momentum and kinetic energy conservation and find v_{1} & v_{2} .$

Using conservation of momentum, initial momentum will be same as final momentum.

$m_{A} v = m_{A} v_{1} + m_{B} v_{2}$

$m_{A} (v - v_{1}) = m_{B} v_{2}$ … $(i)$

Since, this is an elastic collision, the Kinetic Energy shall also remain conserved.

$K . E_{i} = K . E_{f}$

$\frac{1}{2} m_{A} v^{2} = \frac{1}{2} m_{A} v_{1}^{2} + \frac{1}{2} m_{B} v_{2}^{2}$ … $(i i)$

$m_{A} (v^{2} - v_{1}^{2}) = m_{B} v_{2}^{2}$

$m_{A} (v - v_{1}) (v + v_{1}) = m_{B} v_{2}^{2}$

Dividing equation $(i i)$ by $(i)$ , we get,

$\frac{m_{A} (v - v_{1}) (v + v_{1})}{m_{A} (v - v_{1})} = \frac{m_{B} v_{2}^{2}}{m_{B} v^{2}}$

$(v + v_{1}) = v_{2}$

$v = (v_{2} - v_{1})$

Using these values in $(i i)$

$\frac{1}{2} m_{A} (v_{2}^{2} + v_{1}^{2} - 2 v_{1} v_{2}) = \frac{1}{2} m_{A} v_{1}^{2} + \frac{1}{2} m_{B} v_{2}^{2}$

$m_{A} v_{2}^{2} + m_{A} v_{1}^{2} - 2 m_{A} v_{1} v_{2} = m_{A} v_{1}^{2} + m_{B} v_{2}^{2}$

$m_{A} v_{2}^{2} - 2 m_{A} v_{1} v_{2} = m_{B} v_{2}^{2}$

$m_{A} v_{2} - 2 m_{A} v_{1} = m_{B} v_{2}$

$m_{A} (v + v_{1}) - 2 m_{A} v_{1} = m_{B} (v + v_{1})$

$v_{1} = \frac{m_{A} - m_{B}}{m_{A} + m_{B}} v$

$v_{2} = v + v_{1}$

$v_{2} = v + \frac{m_{A} - m_{B}}{m_{A} + m_{B}} v$

$v_{2} = \frac{2 m_{A}}{m_{A} + m_{B}} v$

$Find Initial de-Broglie wavelength for particle A .$

Initial de-Broglie wavelength for particle $A (λ_{i})$ , with initial momentum as $p_{i}$

$λ_{i} = \frac{h}{p_{i}} = \frac{h}{m_{A} \times v}$

$Find final de-Broglie wavelength for particle A .$

Final de-Broglie wavelength for particle $A (λ_{f})$ , with final momentum as $p_{f}$

$λ_{f} = \frac{h}{p_{f}} = \frac{h}{m_{A} \times v_{1}} = \frac{h}{m_{A} (\frac{m_{A} - m_{B}}{m_{A} + m_{B}}) v}$

$λ_{f} = \frac{h (m_{A} + m_{B})}{m_{A} (m_{A} - m_{B}) v}$

Change in de-Broglie wavelength $(Δ λ)$

$Δ λ = λ_{f} - λ_{i} = \frac{h (m_{A} + m_{B})}{m_{A} (m_{A} - m_{B}) v} - \frac{h}{m_{A} \times v}$

$Δ λ = \frac{h}{m_{A} v} (\frac{m_{A} + m_{B}}{m_{A} - m_{B}} - 1)$

$Final Answer : \frac{h}{m_{A} v} (\frac{m_{A} + m_{B}}{m_{A} - m_{B}} - 1)$

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Factors Affecting Kinetic Energy

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