This course provides an introduction to Physics. It is a calculus based
course. The course is examined at two levels, with Physics 1A being the
lower of the two levels.
Mechanics: particle kinematics in one dimension, motion in two and
three dimensions, particle dynamics, work and energy, momentum and
collisions.
Thermal physics: temperature, kinetic theory and the ideal gas, heat and
the first law of thermodynamics.
Waves: oscillations, wave motion, sound waves.
这是一份unsw新南威尔斯大学PHYS1121的成功案例
\begin{prob}
(a) Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of
7.50 m/s. (b) Compare the elephant’s momentum with the momentum of a 0.0400-kg
tranquilizer dart fired at a speed of 600 m/s. (c) What is the momentum of the 90.0-
kg hunter running at 7.40 m/s after missing the elephant?
(a) $p_{\mathrm{e}}=m_{\mathrm{e}} v_{e}=2000 \mathrm{~kg} \times 7.50 \mathrm{~m} / \mathrm{s}=1.50 \times 10^{4} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}$
(b) $p_{\mathrm{b}}=m_{\mathrm{b}} v_{\mathrm{b}}=0.0400 \mathrm{~kg} \times 600 \mathrm{~m} / \mathrm{s}=24.0 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}$, so
$$
\frac{p_{\mathrm{c}}}{p_{\mathrm{b}}}=\frac{1.50 \times 10^{4} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}}{24.0 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}}=625
$$
The momentum of the elephant is much larger because the mass of the elephant is much larger.
(c) $p_{\mathrm{b}}=m_{\mathrm{h}} v_{\mathrm{h}}=90.0 \mathrm{~kg} \times 7.40 \mathrm{~m} / \mathrm{s}=6.66 \times 10^{2} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}$
Again, the momentum is smaller than that of the elephant because the mass of the hunter is much smaller.
PHYS1121 COURSE NOTES ：
A cruise ship with a mass of 1.00 10 kg 7 × strikes a pier at a speed of 0.750 m/s. It
comes to rest 6.00 m later, damaging the ship, the pier, and the tugboat captain’s
finances. Calculate the average force exerted on the pier using the concept of impulse.
(Hint: First calculate the time it took to bring the ship to rest.)
Given: $m=1.00 \times 10^{7} \mathrm{~kg}, v_{0}=0.75 \mathrm{~m} / \mathrm{s}, v=0 \mathrm{~m} / \mathrm{s}, \Delta x=6.00 \mathrm{~m}$. Find: net force on the pier. First, we need a way to express the time, $\Delta t$, in terms of known quantities.
Using the equations $\bar{v}=\frac{\Delta x}{\Delta t}$ and $\bar{v}=\frac{v_{0}+v}{2}$ gives:
$$
\Delta x=\bar{v} \Delta t=\frac{1}{2}\left(v+v_{0}\right) \Delta t \text { so that } \Delta t=\frac{2 \Delta x}{v+v_{0}}=\frac{2(6.00 \mathrm{~m})}{(0+0.750) \mathrm{m} / \mathrm{s}}=16.0 \mathrm{~s} .
$$
net $F=\frac{\Delta p}{\Delta t}=\frac{m\left(v-v_{0}\right)}{\Delta t}=\frac{\left(1.00 \times 10^{7} \mathrm{~kg}\right)(0-0750) \mathrm{m} / \mathrm{s}}{16.0 \mathrm{~s}}=-4.69 \times 10^{5} \mathrm{~N}$.
By Newton’s third law, the net force on the pier is $4.69 \times 10^{5} \mathrm{~N}$, in the original direction of the ship.