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Part (a): Find the speed of the third piece

Step Derivation/Formula Reasoning
1 Let [katex] m [/katex] be the mass of each smaller piece, thus the mass of the third piece is [katex] 2.5m [/katex]. According to the problem, the third piece has 2.5 times the mass of each of the other two pieces.
2 [katex] \vec{p}_{\text{total}} = \vec{0} [/katex] The total initial momentum is zero since the coconut was stationary before exploding.
3 Let [katex] \vec{v}_3 [/katex] be the velocity of the third piece and [katex] \theta [/katex] its angle from west towards south. Then,
[katex]
\vec{p}_{\text{total}} = m \vec{v}_S + m \vec{v}_W + 2.5m \vec{v}_3 = \vec{0}
[/katex]
[katex]
\vec{v}_S = 18\, \text{m/s} \, \hat{j} \quad \text{and} \quad \vec{v}_W = -18\, \text{m/s} \, \hat{i}
[/katex]
[katex]
-m \cdot 18 \, \hat{i} + m \cdot 18 \, \hat{j} + 2.5m \vec{v}_3 = \vec{0}
[/katex]
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 [katex]
\vec{v}_3 = \left(\frac{18}{2.5}\right) \hat{i} – \left(\frac{18}{2.5}\right) \hat{j}
[/katex]
[katex]
\vec{v}_3 = 7.2 \hat{i} – 7.2 \hat{j} \; \text{m/s}
[/katex]
Rearrange to find the velocity vector of the third piece. Cancel [katex]m[/katex] and solve for [katex] \vec{v}_3 [/katex].
5 [katex]
\text{Speed of third piece } |\vec{v}_3| = \sqrt{(7.2)^2 + (7.2)^2} = \sqrt{103.68} \approx 10.18 \; \text{m/s}
[/katex]
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 [katex]
\tan(\theta) = \frac{-7.2}{7.2} = -1
[/katex]
[katex]
\theta = \tan^{-1}(-1) = 135^\circ \, \text{(from east counterclockwise)}
[/katex]
The angle [katex] \theta [/katex] is measured from the negative x-axis, thus the piece is moving to the northeast.
2 135 degrees from east or northwest Answer for part (b).

Part (c): Reducing the impact force of the collision for the bystander

Step Derivation/Formula Reasoning
1 [katex] F = \frac{\Delta p}{\Delta t} [/katex] Force experienced by the bystander can be reduced by increasing the impact time [katex]\Delta t[/katex] or reducing the momentum change [katex]\Delta p[/katex].
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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1. 10.18 m/s
2. 135 degrees from east or northwest
3. Use protective measures or barriers that increase the time of impact, and therefore reduce the force imparted by the mass.

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KinematicsForces
[katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex][katex]F = ma[/katex]
[katex]v = v_i + at[/katex][katex]F_g = \frac{G m_1m_2}{r^2}[/katex]
[katex]a = \frac{\Delta v}{\Delta t}[/katex][katex]f = \mu N[/katex]
[katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex]
Circular MotionEnergy
[katex]F_c = \frac{mv^2}{r}[/katex][katex]KE = \frac{1}{2} mv^2[/katex]
[katex]a_c = \frac{v^2}{r}[/katex][katex]PE = mgh[/katex]
[katex]KE_i + PE_i = KE_f + PE_f[/katex]
MomentumTorque and Rotations
[katex]p = m v[/katex][katex]\tau = r \cdot F \cdot \sin(\theta)[/katex]
[katex]J = \Delta p[/katex][katex]I = \sum mr^2[/katex]
[katex]p_i = p_f[/katex][katex]L = I \cdot \omega[/katex]
Simple Harmonic Motion
[katex]F = -k x[/katex]
[katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex]
[katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex]
ConstantDescription
[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
VariableSI 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]
VariableDerived 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]\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]

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters.

1. Start with the given measurement: [katex]\text{5 km}[/katex]

2. 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]

3. 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]

4. 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]

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity [katex] v_i [/katex] is written as [katex] u [/katex]; sometimes [katex] \Delta x [/katex] is written as [katex] s [/katex].
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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