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. |
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 | 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. |
Phy can also check your working. Just snap a picture!
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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_1m_2}{r^2} |
a = \frac{\Delta v}{\Delta t} | f = \mu N |
R = \frac{v_i^2 \sin(2\theta)}{g} |
Circular Motion | Energy |
---|---|
F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |
a_c = \frac{v^2}{r} | PE = mgh |
KE_i + PE_i = KE_f + PE_f |
Momentum | Torque and Rotations |
---|---|
p = m v | \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 |
---|
F = -k x |
T = 2\pi \sqrt{\frac{l}{g}} |
T = 2\pi \sqrt{\frac{m}{k}} |
Constant | Description |
---|---|
g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
\mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |
k | Spring constant, in \text{N/m} |
M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |
M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |
M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |
Variable | SI Unit |
---|---|
s (Displacement) | \text{meters (m)} |
v (Velocity) | \text{meters per second (m/s)} |
a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |
t (Time) | \text{seconds (s)} |
m (Mass) | \text{kilograms (kg)} |
Variable | Derived SI Unit |
---|---|
F (Force) | \text{newtons (N)} |
E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |
P (Power) | \text{watts (W)} |
p (Momentum) | \text{kilogram meters per second (kgm/s)} |
\omega (Angular Velocity) | \text{radians per second (rad/s)} |
\tau (Torque) | \text{newton meters (Nm)} |
I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |
f (Frequency) | \text{hertz (Hz)} |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: \text{5 km}
Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}
Perform the multiplication: \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}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|
Pico- | p | 10^{-12} | 0.000000000001 |
Nano- | n | 10^{-9} | 0.000000001 |
Micro- | µ | 10^{-6} | 0.000001 |
Milli- | m | 10^{-3} | 0.001 |
Centi- | c | 10^{-2} | 0.01 |
Deci- | d | 10^{-1} | 0.1 |
(Base unit) | – | 10^{0} | 1 |
Deca- or Deka- | da | 10^{1} | 10 |
Hecto- | h | 10^{2} | 100 |
Kilo- | k | 10^{3} | 1,000 |
Mega- | M | 10^{6} | 1,000,000 |
Giga- | G | 10^{9} | 1,000,000,000 |
Tera- | T | 10^{12} | 1,000,000,000,000 |
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