A particle of mass 8 mu g in motion collides | vedclass

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(D) Let the mass of the first particle be $m_1 = 8 \mu g$ and the mass of the second particle be $m_2 = 4 \mu g$. Let the initial velocity of $m_1$ be

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$2$

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(D) Let the mass of the first particle be $m_1 = 8 \mu g$ and the mass of the second particle be $m_2 = 4 \mu g$. Let the initial velocity of $m_1$ be $u_1$ and $m_2$ be $u_2 = 0$. After a perfectly elastic one-dimensional collision, the final velocities $v_1$ and $v_2$ are given by the standard formulas:$v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1 = \frac{8 - 4}{8 + 4} u_1 = \frac{4}{12} u_1 = \frac{1}{3} u_1$$v_2 = \frac{2m_1}{m_1 + m_2} u_1 = \frac{2(8)}{8 + 4} u_1 = \frac{16}{12} u_1 = \frac{4}{3} u_1$The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.The ratio of wavelengths is $\frac{\lambda_1}{\lambda_2} = \frac{h/m_1v_1}{h/m_2v_2} = \frac{m_2v_2}{m_1v_1}$.Substituting the values: $\frac{\lambda_1}{\lambda_2} = \frac{4 	imes (4/3)u_1}{8 	imes (1/3)u_1} = \frac{16/3}{8/3} = \frac{16}{8} = 2 : 1$.

$A$ particle of mass $8 \mu g$ in motion collides with another stationary particle of mass $4 \mu g$. If the collision is perfectly elastic and one-dimensional, the ratio of their de Broglie wavelengths after collision is (in $2 : 1$)

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