Systems of Particles & Linear Momentum

AP Physics C Mechanics guide to systems of particles, center of mass, impulse-momentum theorem with calculus, and collisions in one and two dimensions.

# Systems of Particles & Linear Momentum — AP Physics C Mechanics

AP Physics C extends momentum to systems of particles with calculus-based analysis. You'll need to calculate center of mass for continuous distributions, apply the impulse-momentum theorem with time-varying forces, and solve complex collision problems.

Key Concepts

Linear Momentum

p=mv\vec{p} = m\vec{v}

Newton's Second Law (Momentum Form)

Fnet=dpdt\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}

For constant mass: F=ma\vec{F} = m\vec{a}.

Impulse (Calculus Form)

J=titfF(t)dt=Δp\vec{J} = \int_{t_i}^{t_f} \vec{F}(t)\,dt = \Delta\vec{p}

Center of Mass

Discrete system: xcm=miximix_{\text{cm}} = \frac{\sum m_i x_i}{\sum m_i}

Continuous body: xcm=1Mxdmx_{\text{cm}} = \frac{1}{M}\int x\,dm

For a uniform rod of length LL: xcm=L/2x_{\text{cm}} = L/2 (from one end).

Motion of the Center of Mass

Fext=Macm\vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}} ptotal=Mvcm\vec{p}_{\text{total}} = M\vec{v}_{\text{cm}}

The center of mass moves as if all mass were concentrated there and all external forces acted on it.

Collisions

  • Elastic: conserve both momentum and KE.
  • Inelastic: conserve momentum only.

2D collisions: apply momentum conservation in both xx and yy components independently.

Rocket Propulsion (Variable Mass)

Thrust: Fthrust=vexhaustdmdtF_{\text{thrust}} = v_{\text{exhaust}} \frac{dm}{dt}

v(t)=v0+vexhaustlnm0m(t)v(t) = v_0 + v_{\text{exhaust}} \ln\frac{m_0}{m(t)}

Worked Example

Problem: A time-varying force F(t)=6t2F(t) = 6t^2 N acts on a 3 kg3\ \text{kg} mass from t=0t = 0 to t=2 st = 2\ \text{s}. If the object starts from rest, find the final velocity.

Solution:

J=026t2dt=2t302=16 N\cdotpsJ = \int_0^2 6t^2\,dt = 2t^3\Big|_0^2 = 16\ \text{N·s}

Δp=J=mvf0\Delta p = J = mv_f - 0 vf=1635.33 m/sv_f = \frac{16}{3} \approx 5.33\ \text{m/s}

Practice Questions

  1. 1. Find the center of mass of a uniform thin rod of length LL and mass MM measured from one end.

    xcm=1M0LxMLdx=1LL22=L2x_{\text{cm}} = \frac{1}{M}\int_0^L x \cdot \frac{M}{L}dx = \frac{1}{L}\cdot\frac{L^2}{2} = \frac{L}{2}.

    2. A 2 kg2\ \text{kg} ball moving at 5 m/s5\ \text{m/s} right collides elastically with a 3 kg3\ \text{kg} ball at rest. Find both final velocities (1D).

    v1f=m1m2m1+m2v1i=15(5)=1 m/sv_{1f} = \frac{m_1 - m_2}{m_1 + m_2}v_{1i} = \frac{-1}{5}(5) = -1\ \text{m/s}. v2f=2m1m1+m2v1i=45(5)=4 m/sv_{2f} = \frac{2m_1}{m_1+m_2}v_{1i} = \frac{4}{5}(5) = 4\ \text{m/s}.

    3. A force F(t)=10etF(t) = 10e^{-t} acts from t=0t = 0 to t=t = \infty on a 5 kg5\ \text{kg} mass. What is the total impulse?

    J=010etdt=10[et]0=10(0(1))=10 N\cdotpsJ = \int_0^\infty 10e^{-t}dt = 10[-e^{-t}]_0^\infty = 10(0 - (-1)) = 10\ \text{N·s}.

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Summary

  • F=dp/dt\vec{F} = d\vec{p}/dt is the most general form of Newton's second law.
  • Impulse: J=Fdt=ΔpJ = \int F\,dt = \Delta p.
  • Center of mass: xcm=1Mxdmx_{\text{cm}} = \frac{1}{M}\int x\,dm for continuous bodies.
  • The center of mass of a system responds to external forces only.

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