Particle A experiences a perfectly elastic collision with a stationary particle B. The particles fly apart symmetrically relative to the initial direction of motion of particle A with angle of divergence $θ (0 < θ < 90) .$ Ratio between the masses of the two bodies $\frac{m_{B}}{m_{A}}$ is $\frac{1}{1 + 2 \cos n θ}$ where ‘n’ is ___

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Detailed Solution

$m_{A} u = (m_{A} V_{A} + m_{B} V_{B}) \cos θ$

$m_{A} V_{A} \sin θ = m_{B} V_{B} \sin θ$

$V_{A} = \frac{u}{2} \sec θ; V_{B} = \frac{m_{A}}{m_{B}} \frac{u}{2 \cos θ}$

$\frac{1}{2} m_{A} V_{A}^{2} + \frac{1}{2} m_{B} V_{B}^{2} = \frac{1}{2} m_{A} u^{2}$

$(m_{A} V_{A}) V_{A} + (m_{B} V_{B}) V_{B} = m_{A} u^{2}$

$m_{A} V_{A} [\frac{4}{2 \cos θ} + \frac{m_{A}}{m_{B}} \frac{u}{2 \cos θ}] = m_{A} u^{2}$

On solving this $\frac{m_{A}}{m_{B}} = 4 \cos^{2} θ - 1 = 1 + 2 \cos 2 θ$

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