AP Physics

Unit 5 - Linear Momentum




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# Part A: Calculate velocity of the bullet + wooden block just after impact

Step Derivation/Formula Reasoning
1 m_1 = 0.0500 \, \text{kg} Mass of the bullet.
2 u_1 = 50.0 \, \text{m/s} Initial velocity of the bullet.
3 m_2 = 0.300 \, \text{kg} Mass of the wooden block.
4 u_2 = 0 \, \text{m/s} Initial velocity of the wooden block (at rest).
5 v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} Formula for the final velocity after a completely inelastic collision, where m_1 and m_2 are masses of the bullet and block, u_1 and u_2 are their initial velocities respectively.
6 v = \frac{(0.0500 \, \text{kg}) (50.0 \, \text{m/s}) + (0.300 \, \text{kg}) (0 \, \text{m/s})}{0.0500 \, \text{kg} + 0.300 \, \text{kg}} Substitute the values of m_1 , m_2 , u_1 , and u_2 into the formula.
7 \mathbf{v = 7.14 \, \text{m/s}} Compute to find the final velocity of the bullet + wooden block system just after the collision.

# Part B: Calculate the vertical distance h reached by the bullet and wooden block

Step Derivation/Formula Reasoning
1 v = 7.14 \, \text{m/s} Initial velocity of the system after impact (from part A).
2 v^2 = u^2 + 2gh Use the kinematic equation relating initial velocity (u), final velocity (v), acceleration due to gravity (g), and height (h). Since the system comes to a stop at the highest point, final velocity v is 0.
3 0 = (7.14 \, \text{m/s})^2 + 2(-9.81 \, \text{m/s}^2)h At the highest point the final velocity is 0, and acceleration due to gravity (a negative value, since it acts downwards) impacts the rising object.
4 h = \frac{(7.14 \, \text{m/s})^2}{2 \times 9.81 \, \text{m/s}^2} Rearrange to solve for h.
5 \mathbf{h = 2.59 \, \text{m}} Compute the vertical height.

# Part C: Compare effects of doubling the bullet’s mass vs. velocity on h

Comparison Analysis Conclusion
Doubling bullet’s velocity h_f \propto v^2 , doubling v would quadruple h (since h_f = \frac{4v^2}{2g} ) Increase in velocity has a significant effect, leading to an increase by a factor of four for h .
Doubling bullet’s mass h_f determined more by v (system velocity) than by mass directly. Increasing mass lowers system velocity, hence slightly lowering h . Little to no increase in h ; could actually decrease due to lower velocity after collision.
None Analysis shows that velocity change impacts height significantly compared to mass change. (ii) Doubling the bullet’s velocity has a greater effect on height than doubling the bullet’s mass.

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(a) 7.1 m/s

(b) 2.6 m

(c) ii

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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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