Mechanics

Chapter 11: Impulse and Momentum

Evan Luo

5 min read

1. Momentum - "How much motion?"

When an object of mass mm is moving with velocity v⃗\vec{v}, we say it carries momentum

p⃗=mv⃗\vec{p} = m\vec{v}
  • a vector (it points the same way as v⃗\vec{v})
  • SI unit → kg⋅m/s\text{kg} \cdot \text{m/s}
  • At rest, v=0⇒p⃗=0⃗v=0 \Rightarrow \vec{p}=\vec{0}

Because it is a vector, we can describe each component separately: px=mvxp_x=mv_x, py=mvyp_y=mv_y, …\dots

2. Systems & Total Momentum

For several objects we add their momenta tip-to-tail:

p⃗tot=p⃗1+p⃗2+⋯+p⃗N\vec{p}_{\text{tot}} = \vec{p}_1 + \vec{p}_2 + \dots + \vec{p}_N

Inside the system, the objects can trade momentum with each other, but the only way the total changes is if something outside pushes or pulls.

3. Impulse - "A push that changes momentum"

An external force acting for a time interval gives the system an impulse

J⃗ext=Δp⃗tot\vec{J}_{\text{ext}} = \Delta\vec{p}_{\text{tot}}

If J⃗ext=0⃗\vec{J}_{\text{ext}}=\vec{0}, the system is isolated and its total momentum cannot change.

4. The Law of Conservation of Momentum

For an isolated system:

p⃗tot, before=p⃗tot, after\vec{p}_{\text{tot, before}} = \vec{p}_{\text{tot, after}}

We usually solve problems in two straight-forward steps:

  1. Before: add all p⃗\vec{p} components
  2. After: write the unknowns (v?v?, direction?) and add again
  3. Set "before = after" for both xx- and yy-components and solve

5. Collisions - Three Flavours

Collision typeMomentumMechanical Energy*Notes
Inelasticconserveddecreasesmost real crashes
Perfectly inelasticconserveddecreasesobjects stick together
Perfectly elasticconservedsame before & afterno energy loss; forces are fully elastic

*"Mechanical energy" = kinetic + potential inside the system

6. Recipe for Momentum Problems

  1. Choose the system (all objects that exert big forces on each other)
  2. Draw a "before & after" picture with velocity arrows
  3. Pick xx- and yy-axes
  4. Write px,before=px,afterp_{x,\text{before}} = p_{x,\text{after}} and py,before=py,afterp_{y,\text{before}} = p_{y,\text{after}}
  5. If the collision is elastic, also write energy conservation
  6. Solve for the unknowns, check units & direction

7. Try-It-Yourself Examples

Below each problem is a hidden, step-by-step solution. Read the question, try it yourself, then click Reveal Answer to compare.

8. Example A - Exploding Coconut (2-D)

A 1.0 kg coconut at rest explodes into three pieces:

  • Piece 1: m1=0.230m_1=0.230 kg, v1=11 m/sv_1=11\text{ m/s} west
  • Piece 2: m2=0.320m_2=0.320 kg, v2=14 m/sv_2=14\text{ m/s} north
  • Piece 3: m3=?m_3=? (the rest)

Find the velocity (magnitude & direction) of piece 3.

Solution outline

  1. System = all three fragments; no external impulse → p⃗tot=0⃗\vec{p}_{\text{tot}}=\vec{0}
  2. Mass of piece 3: m3=1.0−0.230−0.320=0.450m_3 = 1.0 - 0.230 - 0.320 = 0.450 kg
  3. Component equations
px:  0=m1(−11)+m2(0)+m3v3xpy:  0=m1(0)+m2(+14)+m3v3y\begin{aligned} p_x:\;& 0 = m_1(-11) + m_2(0) + m_3 v_{3x} \\ p_y:\;& 0 = m_1(0) + m_2(+14) + m_3 v_{3y} \end{aligned}
  1. Solve
v3x=+5.6 m/s,v3y=−9.9 m/sv_{3x} = +5.6\text{ m/s},\qquad v_{3y} = -9.9\text{ m/s}
  1. Magnitude & direction
v3=11.4 m/s,θ=299∘  (measured counter-clockwise from +x)v_3 = 11.4\text{ m/s},\quad \theta = 299^{\circ}\;(\text{measured counter-clockwise from +x})
Click to reveal

9. Example B - Rear-End Collision (Perfectly Inelastic)

A 1500 kg car is stopped at a light. A 1900 kg truck hits it from behind; they lock together and skid 7.15 m before stopping. Road-tire friction μk=0.65\mu_k=0.65. Find the truck's speed just before impact.

  1. Momentum is conserved during the short collision. Let v0v_0 = truck speed pre-impact, vfv_f = common speed right after.
mtv0=(mt+mc)vfm_t v_0 = (m_t+m_c) v_f
  1. Kinetic energy is lost to friction after the crash:
12(mt+mc)vf2=(mt+mc)gμkd\frac{1}{2} (m_t+m_c) v_f^2 = (m_t+m_c) g \mu_k d
  1. Solve the energy equation for vfv_f, then plug into momentum equation for v0v_0.
vf=2gμkd=2(9.8)(0.65)(7.15)≈5.1 m/sv_f = \sqrt{2 g \mu_k d}= \sqrt{2(9.8)(0.65)(7.15)}\approx5.1\text{ m/s}v0=(mt+mc)mtvf≈34001900(5.1)≈17 m/sv_0 = \frac{(m_t+m_c)}{m_t} v_f \approx \frac{3400}{1900}(5.1) \approx 17\text{ m/s}
Click to reveal

10. Example C - Glancing Asteroids (Inelastic, 2-D)

Two 5000 kg asteroids collide: A has vA=40 m/sv_A=40\text{ m/s}, B is at rest. After impact they move apart at angles. Find their speeds and % energy lost.

Follow the "before & after" vector method in both axes; then compare total kinetic energies. You should get vA ⁣f=29.3 m/sv_{A\!f}=29.3\text{ m/s}, vB ⁣f=20.7 m/sv_{B\!f}=20.7\text{ m/s} and about 19.5% of the initial EkE_k is lost.

Click to reveal

11. Example D - Elastic Marble Collision (1-D)

On a friction-free track, a 10 g marble (A) moving left at 0.400 m/s hits a 30 g marble (B) moving right at 0.200 m/s. The impact is perfectly elastic. Find each marble's velocity after.

Because the collision is 1-D and elastic, use both momentum and energy conservation (or the shortcut of "exchange velocities" when masses are equal - but here they're not).

mAvAi+mBvBi=mAvAf+mBvBf12mAvAi2+12mBvBi2=12mAvAf2+12mBvBf2\begin{aligned} m_A v_{A i} + m_B v_{B i} &= m_A v_{A f} + m_B v_{B f} \\ \frac{1}{2} m_A v_{A i}^2 + \frac{1}{2} m_B v_{B i}^2 &= \frac{1}{2} m_A v_{A f}^2 + \frac{1}{2} m_B v_{B f}^2 \end{aligned}

Solving the two equations gives

vAf=−0.10 m/s  (right),vBf=+0.50 m/s  (right)v_{A f} = -0.10\text{ m/s} \;(\text{right}),\quad v_{B f} = +0.50\text{ m/s} \;(\text{right})
Click to reveal

12. Key Takeaways

  • Momentum is the "mass-in-motion" vector
  • Impulse is the "external shove" that changes total momentum
  • If a system is isolated, its total momentum never changes
  • In collisions, momentum is always conserved; energy may or may not be
  • Work problems systematically: choose a system, write component conservation equations, and (if needed) add energy conservation for elastic cases
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