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
$\displaystyle \int_0^3 4t\,dt = 2t^2\big|_0^3 = 18$ N·s.
No external horizontal force; $\vec v_{cm}$ stays zero (ignoring gravity's vertical effect).
Equal masses exchange velocities in a 1D elastic collision.
$p_x = 6.0$ kg·m/s divided by 3.0 kg.
$\displaystyle \frac{3(0) + 1(4)}{4} = 1.0$ m.
$\displaystyle F = v\frac{dm}{dt}$.
The collision is perfectly inelastic.
$\displaystyle \vec F_{ext} = \frac{d\vec p}{dt} = 0$.
Free response
Write your answer first, then open the scoring guide.
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.
Check off the points you earned:
(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.
Check off the points you earned:
(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.
Check off the points you earned: