Newton's second law: $\vec{F} = \frac{d\vec{p}}{dt}$

For constant mass: $\vec{F} = m\frac{d\vec{v}}{dt} = m\vec{a}$

Conservation of Momentum

When $\vec{F}_{ext} = 0$:

$\frac{d\vec{p}_{total}}{dt} = 0$

$\vec{p}_{total} = \text{constant}$

For a system of particles: $\vec{p}_{total} = \sum_i m_i\vec{v}_i = \text{constant}$

Impulse

$\vec{J} = \int_{t_1}^{t_2} \vec{F} \, dt = \Delta \vec{p}$

Variable Force Example

If $F(t) = F_0\sin(\omega t)$ from $t = 0$ to $t = \pi/\omega$:

$J = \int_0^{\pi/\omega} F_0\sin(\omega t) \, dt = \frac{F_0}{\omega}[-\cos(\omega t)]_0^{\pi/\omega}$

$J = \frac{F_0}{\omega}(1 - (-1)) = \frac{2F_0}{\omega}$

Elastic Collisions (1D)

Conservation of momentum: $m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

Conservation of kinetic energy: $\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$

Solutions: $v_{1f} = \frac{(m_1 - m_2)v_{1i} + 2m_2v_{2i}}{m_1 + m_2}$

$v_{2f} = \frac{(m_2 - m_1)v_{2i} + 2m_1v_{1i}}{m_1 + m_2}$

Special Cases

Equal masses ($m_1 = m_2$): $v_{1f} = v_{2i}, \quad v_{2f} = v_{1i}$ (velocities exchange)

Target at rest ($v_{2i} = 0$): $v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}v_{1i}, \quad v_{2f} = \frac{2m_1}{m_1 + m_2}v_{1i}$

Massive target ($m_2 \gg m_1$, $v_{2i} = 0$): $v_{1f} \approx -v_{1i}, \quad v_{2f} \approx 0$ (light object bounces back)

Inelastic Collisions

Momentum conserved, but kinetic energy not conserved.

Perfectly inelastic (objects stick together): $m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_f$

$v_f = \frac{m_1v_{1i} + m_2v_{2i}}{m_1 + m_2}$

Energy lost: $\Delta KE = KE_f - KE_i = \frac{1}{2}\frac{m_1m_2}{m_1 + m_2}(v_{1i} - v_{2i})^2$

(This energy is converted to heat, deformation, sound, etc.)

Coefficient of Restitution

For partially elastic collisions: $e = \frac{|v_{2f} - v_{1f}|}{|v_{1i} - v_{2i}|}$

• $e = 1$: perfectly elastic
• $e = 0$: perfectly inelastic
• $0 < e < 1$: partially elastic

Relative velocity after collision: $v_{2f} - v_{1f} = -e(v_{2i} - v_{1i})$

2D Collisions

Momentum conserved in each direction:

x-component: $m_1v_{1ix} + m_2v_{2ix} = m_1v_{1fx} + m_2v_{2fx}$

y-component: $m_1v_{1iy} + m_2v_{2iy} = m_1v_{1fy} + m_2v_{2fy}$

For elastic collisions, also conserve kinetic energy.

Rocket Motion

Variable mass: $m(t)$ decreases as fuel burns.

$m\frac{dv}{dt} = -v_{rel}\frac{dm}{dt} + F_{ext}$

where $v_{rel}$ is exhaust speed relative to rocket.

In space ($F_{ext} = 0$): $m \, dv = -v_{rel} \, dm$

$\int_{v_0}^v dv' = -v_{rel}\int_{m_0}^m \frac{dm'}{m'}$

$\Delta v = v_{rel}\ln\frac{m_0}{m}$

Tsiolkovsky equation

$v_{1f} = \frac{-1.0}{5.0}(5.0) = \boxed{-1.0 \text{ m/s}}$

(Block 1 bounces backward)

$v_{2f} = \frac{2m_1}{m_1 + m_2}v_{1i} = \frac{2(2.0)}{5.0}(5.0)$

$v_{2f} = \frac{4.0}{5.0}(5.0) = \boxed{4.0 \text{ m/s}}$

(b) Momentum conservation check:

Initial: $p_i = (2.0)(5.0) + 0 = 10$ kg·m/s

Final: $p_f = (2.0)(-1.0) + (3.0)(4.0) = -2 + 12 = 10$ kg·m/s ✓

(c) Kinetic energy conservation check:

Initial: $KE_i = \frac{1}{2}(2.0)(5.0)^2 = 25 \text{ J}$

Final: $KE_f = \frac{1}{2}(2.0)(-1.0)^2 + \frac{1}{2}(3.0)(4.0)^2$

$KE_f = 1 + 24 = 25 \text{ J}$ ✓

$h = 1.5(1 - 0.766) = 1.5(0.234) = 0.351 \text{ m}$

Energy conservation (after collision): $\frac{1}{2}(M + m)v_1^2 = (M + m)gh$

$v_1 = \sqrt{2gh} = \sqrt{2(9.8)(0.351)}$

$\boxed{v_1 = 2.62 \text{ m/s}}$

(b) Initial bullet velocity:

Momentum conservation (during collision): $mv_0 = (M + m)v_1$

$v_0 = \frac{(M + m)v_1}{m} = \frac{(2.01)(2.62)}{0.01}$

$\boxed{v_0 = 527 \text{ m/s}}$

(c) Energy lost percentage:

Initial KE: $KE_i = \frac{1}{2}mv_0^2 = \frac{1}{2}(0.01)(527)^2 = 1389 \text{ J}$

KE just after collision: $KE_f = \frac{1}{2}(M + m)v_1^2 = \frac{1}{2}(2.01)(2.62)^2 = 6.90 \text{ J}$

Percentage lost: $\frac{KE_i - KE_f}{KE_i} \times 100\% = \frac{1389 - 6.90}{1389} \times 100\%$

$\boxed{99.5\% \text{ of energy lost}}$

Most energy goes into deformation, heat, and sound!