AP Physics Unit

Unit 5 - Linear Momentum




A 0.0350 kg bullet moving horizontally at 425 m/s embeds itself into an initially stationary 0.550 kg block.

  1. (a) What is their velocity just after the collision? (1 points)
  2. (b) The bullet-embedded block slides on horizontal surface with a 0.40 kinetic coefficient of friction. What is its velocity after sliding 10.0 meters? (3 points)
  3. (c) The bullet-embedded block then strikes and sticks to a stationary 2.50 kg block (at the 10.0 m point). How far does system travel before stopping? (3 points)

Velocity just after the collision

Step Formula Derivation Reasoning
1 m_1v_1 + m_2v_2 = (m_1 + m_2)v’ Conservation of momentum, where m_1 and m_2 are the masses of the bullet and block, v_1 and v_2 are their initial velocities, and v’ is their final velocity.
2 v’ = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} Solve for v’. Given: m_1 = 0.0350 , \text{kg}, v_1 = 425 , \text{m/s}, m_2 = 0.550 , \text{kg}, v_2 = 0 , \text{m/s}.
3 v’ = 25.43 , \text{m/s} The velocity just after the collision
Velocity after Sliding 10.0 m
Step Formula Derivation Reasoning
1 v^2 = v’^2 + 2ad Kinematic equation for motion under constant acceleration, where v is the final velocity, v’ is the initial velocity, a is acceleration, and d is the distance.
2 a = -\mu_k g Acceleration due to kinetic friction, where \mu_k is the coefficient of kinetic friction and g is the acceleration due to gravity (approximately 9.81 , \text{m/s}^2). Given: \mu_k = 0.40.
3 v = \sqrt{v’^2 + 2ad} Solve for v. Given: v’ = 25.43 , \text{m/s}, d = 10.0 , \text{m}.
4 v = 23.84 , \text{m/s} Velocity after sliding 10.0 meters


Distance traveled by the combined system (2 blocks and the bullet)

Step Formula Derivation Reasoning
1 m_1v_1 + m_2v_2 = (m_1 + m_2)v’ Conservation of momentum for the collision between the bullet-block system and the second block, where m_1 and v_1 are the mass and velocity of the bullet-block system, m_2 and v_2 are the mass and velocity of the second block, and v’ is the final velocity of the combined system.
2 v’ = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} Solve for v’. Given: m_1 = 0.585 , \text{kg}, v_1 = 23.84 , \text{m/s}, m_2 = 2.50 , \text{kg}, v_2 = 0 , \text{m/s}.
3 0 = v’^2 + 2ad Kinematic equation for motion under constant acceleration when the final velocity is 0.
4 d = \frac{-v’^2}{2a} Solve for d. The acceleration a remains -\mu_k g as before.
5 v’ = 4.52 , \text{m/s} Final velocity of the combined system after the second collision
6 d = 2.60 , \text{m} Distance traveled by the combined system before stopping.
  1. 25.43 m/s
  2. 23.84 m/s
  3. 2.6 m

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  1. 25.43 m/s
  2. 23.84 m/s
  3. 2.6 m

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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




















(Base unit)


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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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