Source: High school physics (Chinese)
Problem Sets:
Problem
A proton (mass $m_p = 1.67 \times 10^{-27}$ kg) moving at $v_p = 1.0 \times 10^7$ m/s collides with a nitrogen nucleus initially at rest ($v_N = 0$). After the collision, the proton rebounds at $v_p' = -6.0 \times 10^6$ m/s, and the nitrogen nucleus moves forward at $v_N' = 4.0 \times 10^6$ m/s. The velocities are along a straight line.
- What is the mass of the nitrogen nucleus?
- Can you find the interaction force during their collision? Why?
[Q1] $m_N = 6.68 \times 10^{-27}$ kg [Q2] No.
[Q1] We apply the principle of conservation of linear momentum to the system. Let the initial direction of the proton be positive.
$$P_{initial} = P_{final}$$ $$m_p v_p + m_N v_N = m_p v_p' + m_N v_N'$$Since $v_N=0$, we can solve for the mass of the nitrogen nucleus, $m_N$:
$$m_N = \frac{m_p(v_p - v_p')}{v_N'}$$Substituting the given values:
$$m_N = \frac{(1.67 \times 10^{-27} kg)(1.0 \times 10^7 m/s - (-6.0 \times 10^6 m/s))}{4.0 \times 10^6 m/s}$$ $$m_N = \frac{(1.67 \times 10^{-27} kg)(1.6 \times 10^7 m/s)}{4.0 \times 10^6 m/s} = 6.68 \times 10^{-27} kg$$[Q2] The interaction force cannot be determined. The average force is related to the change in momentum (impulse) by $F_{avg} = \Delta p / \Delta t$. While the change in momentum $\Delta p$ can be calculated, the duration of the collision, $\Delta t$, is unknown. Without $\Delta t$, the force cannot be found.