AP Physics C: Mechanics

Unit 4

Linear Momentum

8 practice questions · Free response

Notes

Momentum and impulse

  • $\vec p = m\vec v$ and $\displaystyle \vec F_{net} = \frac{d\vec p}{dt}$.
  • Impulse: $\displaystyle \vec J = \int \vec F\,dt = \Delta\vec p$. If $F(t)$ is given, integrate it.

Systems and the center of mass

  • $\displaystyle \vec v_{cm} = \frac{\sum m_i\vec v_i}{\sum m_i} = \frac{\vec p_{total}}{M}$.
  • With no net external force, the center of mass moves at constant velocity, even during an explosion.
  • $\vec F_{ext} = M\vec a_{cm}$.

Collisions

  • Momentum is conserved in every collision if external forces are negligible during the short impact.
  • Elastic: kinetic energy is also conserved. In a 1D elastic collision, the relative speed of approach equals the relative speed of separation.
  • Perfectly inelastic: objects stick: $\displaystyle v_f = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}$.
  • 2D collisions: conserve momentum separately in $x$ and in $y$.

Variable mass (rockets, sand dropping on a belt)

$\displaystyle F = \frac{dp}{dt} = m\frac{dv}{dt} + v\frac{dm}{dt}$. A conveyor belt at constant speed $v$ collecting sand at rate $\displaystyle \frac{dm}{dt}$ needs extra force $\displaystyle v\frac{dm}{dt}$.

Practice

A force $F(t) = 4t$ N acts on a 2.0 kg object from $t = 0$ to $t = 3$ s. What impulse does it deliver?
A shell at rest explodes into two pieces. The center of mass of the pieces:
A 1.0 kg ball at 4.0 m/s collides elastically head-on with an identical ball at rest. After the collision:
A 2.0 kg puck moving east at 3.0 m/s sticks to a 1.0 kg puck moving north at 3.0 m/s. What is the final velocity's east component?
A 3.0 kg and a 1.0 kg object are at $x = 0$ and $x = 4.0$ m. Where is their center of mass?
Sand falls onto a conveyor belt at 5.0 kg/s, and the belt moves at a constant 2.0 m/s. What extra force is needed to keep the belt moving?
In a ballistic pendulum, during the brief collision between the bullet and the block, which is conserved?
The net external force on a system is zero. Which must be true?

Free response

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Free-response question

A 0.010 kg bullet moving at $v_0$ embeds in a 1.99 kg block hanging from a string. The block swings up 0.20 m. Use $g = 10$ m/s².

(a) Find the speed of the block and bullet right after the collision.

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$\displaystyle \frac{1}{2}Mv^2 = Mgh \Rightarrow v = \sqrt{2(10)(0.20)} = 2.0$ m/s.

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(b) Find the bullet's initial speed $v_0$.

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$0.010v_0 = 2.00(2.0) \Rightarrow v_0 = 400$ m/s.

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(c) Explain why you can't use energy conservation from the moment before impact to the top of the swing.

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The collision is perfectly inelastic: most of the bullet's kinetic energy turns into internal energy as it embeds, so mechanical energy isn't conserved across the collision.

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