Part (a): Find the speed of the third piece
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | Let \( m \) be the mass of each smaller piece, thus the mass of the third piece is \( 2.5m \). | According to the problem, the third piece has 2.5 times the mass of each of the other two pieces. |
| 2 | \( \vec{p}_{\text{total}} = \vec{0} \) | The total initial momentum is zero since the coconut was stationary before exploding. |
| 3 | Let \( \vec{v}_3 \) be the velocity of the third piece and \( \theta \) its angle from west towards south. Then, \( \vec{p}_{\text{total}} = m \vec{v}_S + m \vec{v}_W + 2.5m \vec{v}_3 = \vec{0} \) \( \vec{v}_S = 18\, \text{m/s} \, \hat{j} \quad \text{and} \quad \vec{v}_W = -18\, \text{m/s} \, \hat{i} \) \( -m \cdot 18 \, \hat{i} + m \cdot 18 \, \hat{j} + 2.5m \vec{v}_3 = \vec{0} \) |
Set the total momentum as the vector sum of individual momenta. The pieces are moving south and west with the same speed but in perpendicular directions. |
| 4 | \( \vec{v}_3 = \left(\frac{18}{2.5}\right) \hat{i} – \left(\frac{18}{2.5}\right) \hat{j} \) \( \vec{v}_3 = 7.2 \hat{i} – 7.2 \hat{j} \; \text{m/s} \) |
Rearrange to find the velocity vector of the third piece. Cancel \(m\) and solve for \( \vec{v}_3 \). |
| 5 | \( \text{Speed of third piece } |\vec{v}_3| = \sqrt{(7.2)^2 + (7.2)^2} = \sqrt{103.68} \approx 10.18 \; \text{m/s} \) |
Calculate the magnitude to find the speed of the third piece. |
| 6 | 10.18 m/s | Answer for part (a), the speed of the third piece. |
Part (b): Find the direction of the third piece
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( \tan(\theta) = \frac{-7.2}{7.2} = -1 \) \( \theta = \tan^{-1}(-1) = 135^\circ \, \text{(from east counterclockwise)} \) |
The angle \( \theta \) is measured from the negative x-axis, thus the piece is moving to the northeast. Alternatively we can state that its 45 degrees north east. |
| 2 | 45 degrees north east | Answer for part (b). |
Part (c): Reducing the impact force of the collision for the bystander
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( F = \frac{\Delta p}{\Delta t} \) | Force experienced by the bystander can be reduced by increasing the impact time \(\Delta t\) or reducing the momentum change \(\Delta p\). |
| 2 | Wear protective gear or position a net/barrier | By wearing protective gear or positioning an absorbent barrier (like a net), the bystander can prolong the impact time and reduce the force. |
| 3 | Use protective measures or barriers | Answer for part (c), suggesting that protective gear or an impact-absorbing barrier would reduce the effect of the collision. |
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Two people, one of mass \( 88 \) \( \text{kg} \) and the other of mass \( 55 \) \( \text{kg} \), sit in a rowboat of mass \( 70 \) \( \text{kg} \). With the boat initially at rest, the two people, who have been sitting at opposite ends of the boat \( 3.1 \) \( \text{m} \) apart from each other, now exchange seats.
Two ice skaters suddenly push off against one another starting from a stationary position. The \(45 \, \text{kg}\) skater acquires a speed of \(0.375 \, \text{m/s}\) relative to the ice. What speed does the \(60 \, \text{kg}\) skater acquire relative to the ice?
Astronaut Jennifer’s lifeline to her spaceship comes loose and she finds herself stranded, “floating” \( 100 \) \( \text{m} \) from the mothership. She suddenly throws her \( 2.00 \) \( \text{kg} \) wrench at \( 20 \) \( \text{m/s} \) in a direction away from the ship. If she and her spacesuit have a combined mass of \( 200 \) \( \text{kg} \), how long does it take her to coast back to her spaceship?
A \(1200 \, \text{kg}\) car moving at \(15.6 \, \text{m/s}\) suddenly collides with a stationary car of mass \(1500 \, \text{kg}\). If the two vehicles lock together, what is their combined velocity immediately after the collision?
A platform is initially rotating on smooth ice with negligible friction, as shown above. A stationary disk is dropped directly onto the center of the platform. A short time later, the disk and platform rotate together at the same angular velocity, as shown at right in the figure. How does the angular momentum of only the platform change, if at all, after the disk drops? And what is the best justification.
A \(2,000 \, \text{kg}\) car collides with a stationary \(1,000 \, \text{kg}\) car. Afterwards, they slide \(6 \, \text{m}\) before coming to a stop. The coefficient of friction between the tires and the road is \(0.7\). Find the initial velocity of the \(2,000 \, \text{kg}\) car before the collision?
A \(15 \, \text{g}\) marble moves to the right at \(3.5 \, \text{m/s}\) and makes an elastic head-on collision with a \(22 \, \text{g}\) marble. The final velocity of the \(22 \, \text{g}\) marble is \(2.0 \, \text{m/s}\) to the right, and the final velocity of the \(15 \, \text{g}\) marble is \(5.4 \, \text{m/s}\) to the left. What was the initial velocity of the \(22 \, \text{g}\) marble?
A 0.035 kg bullet moving horizontally at 350 m/s embeds itself into an initially stationary 0.55 kg block. Air resistance is negligible.
A \( 1.0 \)\( \text{-kg} \) object is moving with a velocity of \( 6.0 \) \( \text{m/s} \) to the right. It collides and sticks to a \( 2.0 \)\( \text{-kg} \) object moving with a velocity of \( 3.0 \) \( \text{m/s} \) in the same direction. How much kinetic energy was lost in the collision?
A rubber ball with a mass of \(0.25 \, \text{kg}\) and a speed of \(19.0 \, \text{m/s}\) collides perpendicularly with a wall and bounces off with a speed of \(21 \, \text{m/s}\) in the opposite direction. What is the magnitude of the impulse acting on the rubber ball?
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
Metric Prefixes
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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