A particle A of mass m and initial velocity v | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the velocity of particle $A$ after the collision be $v_{1}$ and the velocity of particle $B$ be $v_{2}$.Using the law of conservation of linea

Vedclass · Accepted Answer

$\frac{\lambda_{A}}{\lambda_{B}}=2$

Vedclass · Answer

(C) Let the velocity of particle $A$ after the collision be $v_{1}$ and the velocity of particle $B$ be $v_{2}$.Using the law of conservation of linear momentum:$m v = m v_{1} + (m/2) v_{2}$$2v = 2v_{1} + v_{2} \quad \dots (i)$For an elastic head-on collision,the coefficient of restitution $e = 1$:$e = \frac{v_{2} - v_{1}}{v - 0} = 1$$v = v_{2} - v_{1} \implies v_{2} = v + v_{1} \quad \dots (ii)$Substituting $(ii)$ into $(i)$:$2v = 2v_{1} + (v + v_{1})$$v = 3v_{1} \implies v_{1} = \frac{v}{3}$$v_{2} = v + \frac{v}{3} = \frac{4v}{3}$The de-Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{mv}$.Ratio of wavelengths: $\frac{\lambda_{A}}{\lambda_{B}} = \frac{p_{B}}{p_{A}} = \frac{(m/2) v_{2}}{m v_{1}}$$\frac{\lambda_{A}}{\lambda_{B}} = \frac{(m/2) 	imes (4v/3)}{m 	imes (v/3)} = \frac{2mv/3}{mv/3} = 2$.

$A$ particle $A$ of mass $m$ and initial velocity $v$ collides with a particle $B$ of mass $\frac{m}{2}$ which is at rest. The collision is head-on and elastic. The ratio of de-Broglie wavelengths $\lambda_{A}$ to $\lambda_{B}$ after the collision is

Similar Questions

In adiabatic compression,the decrease in volume is associated with

Differentiate the function with respect to $x$: $\sec (\tan (\sqrt{x}))$

The radius of an atom is of the order of

In the structure of $B_2H_6$,the number of $BH_2$ groups present in one plane,and the number of $B-H$ terminal bonds,$B-B$ bonds,and $B-H-B$ bridge bonds are respectively:

$C_2H_6 \stackrel{450^{\circ}C}{\longrightarrow} C_2H_4 + H_2$. The above reaction is called:

Vedclass Test Series

Exam Paper Generator

Online Exam Module