Newton's third law states that for every action there is an equal and opposite reaction. Consider two objects \(O_1\) (mass \(m_1\)) and \(O_2\) (mass \(m_2\)) moving toward each other on a frictionless surface and colliding. Let \(F_{12}\) be the force exerted by \(O_1\) on \(O_2\) and \(F_{21}\) the force exerted by \(O_2\) on \(O_1\). Then: \[ F_{12} = -F_{21} \] Using Newton's second law, \(F = m a\), and acceleration \(a = \frac{\Delta v}{\Delta t}\): \[ m_1 \left( \frac{\Delta v}{\Delta t} \right)_{O_1} = - m_2 \left( \frac{\Delta v}{\Delta t} \right)_{O_2} \] \[ m_1 (\Delta v)_{O_1} = - m_2 (\Delta v)_{O_2} \] Let \(v_{1i}\) and \(v_{1f}\) be the initial and final velocities of \(O_1\), and \(v_{2i}\) and \(v_{2f}\) for \(O_2\): \[ m_1 (v_{1f} - v_{1i}) = - m_2 (v_{2f} - v_{2i}) \] Rearranged: \[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \]
Momentum before interaction = Momentum after interaction
Thus, with no external forces, momentum is conserved.A 35 kg boy jumps from rest into a moving trolley of mass 70 kg already moving at 5 m/s to the right. What is the trolley's speed after?
Solution
Initial velocities: \(v_{1i} = 0\), \(v_{2i} = 5 \text{ m/s}\)
Initial momentum:
\[
p_i = m_1 v_{1i} + m_2 v_{2i} = 35 \times 0 + 70 \times 5 = 350 \; \mathrm{kg\cdot m/s}
\]
Let \(v\) be the final combined velocity. Final momentum:
\[
p_f = (m_1 + m_2) v = (35 + 70) v
\]
Conservation of momentum:
\[
350 = 105 v
\]
\[
v = \frac{350}{105} \approx 3.3 \text{ m/s (right)}
\]
A 35 kg boy running at 2 m/s to the right jumps onto a stationary 70 kg trolley. What is the combined speed?
Solution
Initial velocities: \(v_{1i} = 2 \text{ m/s}\), \(v_{2i} = 0\)
Initial momentum:
\[
p_i = 35 \times 2 + 70 \times 0 = 70 \; \mathrm{kg\cdot m/s}
\]
Let \(v\) be the final combined velocity:
\[
p_f = (35 + 70) v = 105 v
\]
Conservation:
\[
70 = 105 v
\]
\[
v = \frac{70}{105} \approx 0.7 \text{ m/s (right)}
\]
A 5 kg gun fires a 15 g bullet at 1000 m/s to the right. What is the gun's recoil velocity?
Solution
Initial velocities: \(v_{1i} = 0\), \(v_{2i} = 0\)
Initial momentum: \(p_i = 0\)
Final velocities: let \(v_{1f}\) be the gun's recoil velocity, \(v_{2f} = 1000 \text{ m/s}\).
Final momentum:
\[
p_f = 5 v_{1f} + 0.015 \times 1000 = 5 v_{1f} + 15
\]
Conservation:
\[
0 = 5 v_{1f} + 15
\]
\[
v_{1f} = -\frac{15}{5} = -3 \text{ m/s (left)}
\]
1. Higher Level Physics - IB Diploma - Chris Hamper, Pearson
2. Physics - Raymond A. Serway and Jerry S. Faughn, Holt, Rinehart and Winston