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) 5/6
B) 11/12
C) 1/12
D) 10/13
E) 11/13

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, $v$ is the initial velocity, $m_1$ is the mass of the moving particle (neutron), and $m_2$ is the mass of the stationary particle (carbon nucleus). 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, yielding $v' = \frac{(1m - 12m)}{(1m + 12m)}v = \frac{-11m}{13m}v$. The negative sign indicates direction change, but for the fraction of its initial speed, we take the absolute value, which is $\frac{11}{13}$.

In a Certain Nuclear Reactor, Neutrons Suddenly Collide with Carbon