AP Physics

Unit 5 - Linear Momentum




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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.

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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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\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_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 MotionEnergy
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
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients 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.
kSpring 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
VariableSI 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)}
VariableDerived 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. 

  1. Start with the given measurement: \text{5 km}

  2. 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}}

  3. 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}

  4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}



Power of Ten




















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  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 v_i is written as u ; sometimes \Delta x is written as s .
  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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