A proton of mass m collides elastically with | vedclass

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

Question

(D) Let the initial velocity of the proton be $u$ and the final velocities of the proton and the particle be $v_1$ and $v_2$ respectively. Let the pro

Vedclass · Accepted Answer

$m$

Vedclass · Answer

(D) Let the initial velocity of the proton be $u$ and the final velocities of the proton and the particle be $v_1$ and $v_2$ respectively. Let the proton move at an angle $	heta$ with the initial direction, then the particle moves at an angle $(90^\circ - 	heta)$.By conservation of linear momentum:Along the initial direction ($x$-axis): $mu = mv_1 \cos 	heta + Mv_2 \cos(90^\circ - 	heta) = mv_1 \cos 	heta + Mv_2 \sin 	heta$ ...$(i)$Along the perpendicular direction ($y$-axis): $0 = mv_1 \sin 	heta - Mv_2 \sin(90^\circ - 	heta) = mv_1 \sin 	heta - Mv_2 \cos 	heta$ ...$(ii)$From $(ii)$, $Mv_2 \cos 	heta = mv_1 \sin 	heta$, so $Mv_2 = mv_1 	an 	heta$.Since the collision is elastic, kinetic energy is conserved:$\frac{1}{2}mu^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}Mv_2^2$$mu^2 = mv_1^2 + Mv_2^2$ ...$(iii)$Squaring and adding equations $(i)$ and $(ii)$:$(mu)^2 = (mv_1 \cos 	heta + Mv_2 \sin 	heta)^2 + (mv_1 \sin 	heta - Mv_2 \cos 	heta)^2$$m^2u^2 = m^2v_1^2 + M^2v_2^2$ ...$(iv)$Comparing $(iii)$ and $(iv)$:$m(mv_1^2 + Mv_2^2) = m^2v_1^2 + M^2v_2^2$$m^2v_1^2 + m M v_2^2 = m^2v_1^2 + M^2v_2^2$$m M v_2^2 = M^2v_2^2$$m = M$Thus, the mass of the unknown particle is $m$.

$A$ proton of mass $m$ collides elastically with a particle of unknown mass $M$ at rest. After the collision, the proton and the unknown particle are seen moving at an angle of $90^\circ$ with respect to each other. The mass of the unknown particle is:

Similar Questions

$A$ neutron makes a head-on elastic collision with a stationary deuteron. The fractional energy loss of the neutron in the collision is

What happens when two identical bodies moving in opposite directions undergo a perfectly elastic collision?

$A$ neutron travelling with a velocity $v$ and kinetic energy $E$ collides perfectly elastically head-on with the nucleus of an atom of mass number $A$ at rest. The fraction of total energy retained by the neutron is

$A$ particle of mass $m$ moving with kinetic energy $E$ collides head-on elastically with another particle of mass $2m$ at rest. After the collision,the kinetic energy of the first particle is:

For a perfectly elastic collision,the coefficient of restitution $e$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module