An unknown nucleus of mass m_1 collides with | vedclass

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(C) In a collision where a projectile of mass $m_1$ strikes a stationary target of mass $m_2$,and the two particles move at an angle of $90^{\circ}$ t

Vedclass · Accepted Answer

${}^{12}C$

Vedclass · Answer

(C) In a collision where a projectile of mass $m_1$ strikes a stationary target of mass $m_2$,and the two particles move at an angle of $90^{\circ}$ to each other,the condition is given by $m_2 = m_1 \cdot e$,where $e$ is the coefficient of restitution.For an inelastic collision,kinetic energy is lost,which implies $e Substituting $e  m_2$.Since the target nucleus is ${}^4He$ $(m_2 = 4 	ext{ amu})$,the projectile nucleus must have a mass $m_1 > 4 	ext{ amu}$.Among the given options,${}^{28}N$ and ${}^{12}C$ are heavier than ${}^4He$. However,in standard physics problems of this type,the projectile is typically heavier than the target to satisfy the scattering geometry. ${}^{12}C$ is a valid candidate.

An unknown nucleus of mass $m_1$ collides with a stationary ${}^4He$ nucleus of mass $m_2$. After the collision,the two nuclei travel in perpendicular directions relative to each other. If kinetic energy is lost in the collision,the unknown nucleus must be:

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