Momentum conservation and recoil
Internal forces always cancel in pairs, so an isolated system's total momentum never changes — which is why cannons recoil and rockets fly.
When it holds
Every internal force in a system has an equal and opposite partner (Newton's third law), so internal pushes shuffle momentum between parts of the system but never change the total. The total momentum of a system stays constant unless an outside force acts. Friction with the ground, gravity on a projectile, or a wall all count as outside.
m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'Center of mass
Total momentum equals total mass times the center-of-mass velocity. With no external force, the center of mass keeps moving in a straight line at constant speed no matter how violent the internal motion gets — an exploding shell's fragments scatter, but their center of mass traces the same arc the intact shell would have followed.
v_{\text{cm}} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}Worked example
A 1200 kg cannon fires a 15 kg cannonball horizontally at 400 m/s. Total momentum before firing is zero, so after firing it must still be zero: the cannon recoils at (15)(400)/1200 = 5.0 m/s backward. Both objects gained kinetic energy — burning powder paid for it — yet the momentum still sums to zero.
0 = (15)(400) + (1200)v \;\Rightarrow\; v = -5.0\ \text{m/s}Flashcard
Total momentum of an isolated system is fixed
Check your understanding
A 40 kg child standing still on frictionless ice throws a 4.0 kg rock horizontally at 5.0 m/s. What is the child's recoil speed?