In a certain nuclear reactor,neutrons suddenly collide with carbon nuclei,which have 12 times the mass of neutrons.In a head-on elastic collision with a stationary carbon nucleus,what fraction of its initial speed does the neutron have after the collision?
A)11/12
B)1/12
C)11/13
D)5/6
E)10/13

Question

Quizplus · Accepted Answer

In an elastic collision, the velocity of a particle after collision can be determined using the formula: $v' = \frac{(m_1 - m_2)}{(m_1 + m_2)}v$, where $v'$ is the velocity of the particle after collision, $m_1$ is the mass of the moving particle (neutron), $m_2$ is the mass of the stationary particle (carbon nucleus), and $v$ is the initial velocity of the moving particle. Given that the carbon nucleus has 12 times the mass of the neutron, we substitute $m_1 = 1m$ and $m_2 = 12m$ into the formula, resulting in $v' = \frac{(1m - 12m)}{(1m + 12m)}v = \frac{-11m}{13m}v$. The negative sign indicates direction change, but the question asks for the fraction of the initial speed, which is the absolute value, hence $|-\frac{11}{13}|$ or $\frac{11}{13}$.

In a Certain Nuclear Reactor,neutrons Suddenly Collide with Carbon Nuclei,which