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Linear Momentum Speed Review

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Jason Kuma

Writer | Coach | Builder | Fremont, CA

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This article will simplify linear momentum and address many misconceptions. By the end of the article you will be able to solve any problem using conservation of momentum or the impulse-momentum law. Note that this doesn’t cover angular (or rotational) momentum. To learn about that click this link.

What is momentum?

Anything that has mass (which is pretty much everything) AND has velocity has momentum. Hence the equation p = mv.

Conservation of Momentum

Means that the initial momentum value should equal the final. If they are not equal then momentum is NOT conserved and there has been an impulse caused by an external force (more on this later).

Conservation of momentum: <sub> </sub> p_i = p_f

To solve problems add up the momentums’ of objects before a collision. Then set it equal to the sum of momentum of the objects after a collision.

Three Types of Collisions:

  1. Elastic – Objects collide and bounce of each other (energy is generally conserved in this type of collision)
  2. Inelastic – Objects collide and stick to each other.
  3. Explosion – Object starts at rest then explodes outwards in multiple pieces.

Momentum is conserved in ALL collision. Energy, however, might not be conserved. Why? When objects collide a small about of heat is dissipated and lost to the surround environment.

Example of Conservation

Let’s try the following problem:

A 0.450 kg ice puck, moving east with a speed of 3.00 m/s, has a head-on collision with a 0.900 kg puck initially at rest. Assuming a perfectly elastic collision, what will be the speed and direction of each object after the collision?

Step one: Identify if momentum is conserved. This problem deals with a elastic collision, so momentum is definitely conserved. Therefore, we can use conservation of momentum

Step two: Apply conservation of momentum – Pi = Pf –> (.45)(3) = (.9)(v1) + (.45)(v2)

Step three: notice how you have too many unknowns to be able to solve this problem. You need a second equation. This will come from the conservation of energy (we can use this because the collision is elastic). Ei = Ei –> (.45)(3)2 = (.9)(v2)2 + (.45)(v1)2

Step four: rearrange and solve the equations above. Use any method that works best for you. I would recommend substitution in this case.

Step five: final solution, v1 = -1 m/s, v2 = 2 m/s


Impulse is a change in momentum. This means that Pi = Pf.

\text{Impulse} = \Delta p = m \Delta v = Ft

The equations above show: the change in momentum, which is really a change in velocity, comes from an external force being applied. This makes sense since a force implies an acceleration, which is a change in velocity.

What is an external force? It is a force that is not within the system. For example, imagine a car traveling. The system is just the car. Suppose it hits a wall and comes to a stop. The wall applied an external force, that caused a change in velocity, thus a change in momentum.

Example Impulse Problem

A 0.0600kg tennis ball is traveling at 30.0m/s. After being hit by the opponents racket, the ball’s velocity is 20.0 m/s in the opposite direction. Calculate the:
a) change in the ball’s momentum and
b) average force exerted by the racket if the ball and racket were in contact for 0.0400 s


  1. Check if momentum is conserved. It is NOT conserved in this problem, since the racket is applying an external force. Hence, we will use the Impulse theorm to solve this.
  2. Re-arrange and apply Impulse equations –> ∆p = m∆v –> (.06)(30+20) = 3 Ns [make sure to apply signs correctly, as this is one of the most common mistakes]
  3. Solve the part (b) by apply the other equation for impulse. ∆p = Ft –> 3 = F(.04) –> F = 75N

Other Important Concepts for the AP Exam

  1. Velocity of center of mass. As long as momentum is conserved, so will the velocity of the center of mass.
  2. Collisions at angles. Apply the same laws and equations as collisions in one dimension, but this time split velocity into x and y components and apply conservation in both directions.

That’s a wrap! If you are still having difficulty applying these concepts be sure to ask you teachers or friends for guidance. If you want quick and easy guidance, consider hiring a tutor here! Nerd-Notes guarantees a boost in grades and understanding in just the first lesson.

10 Problems to Help You Master Momentum!

If you get all of these right, you have reached mastery. If you score anything below a 70% — you may want to review concepts or problem solving strategies in detail.

Question 1
Difficulty - Advanced
Solve Type - Mathematical
A bullet of mass 0.0500 kg traveling at 50.0 m/s is fired horizontally into a wooden block suspended from a long rope. The mass of the wooden block is 0.300 kg and it is initially at rest. The collision is completely inelastic and after impact the bullet+ wooden block move together until the center of mass of the system rises a vertical distance h above its initial position.
View Full Question and Explanation
Question 2
Difficulty - Beginner
Solve Type - Mathematical
A child (mass 32 kg) in a boat (mass 71 kg) throws a 7.1 kg package out horizontally with a speed of 12.2 m/s. Calculate the velocity of the boat immediately after, assuming it was initially at rest. Ignore water resistance.
View Full Question and Explanation
Question 3
Difficulty - Beginner
Solve Type - Mathematical
A pool cue ball, mass 0.7 kg, is traveling at 2 m/s when it collides head on with another ball, mass 0.5 kg, traveling in the opposite direction with a speed of 1.2 m/s. After the collision, the cue ball travels in the opposite direction at 0.3 m/s. What is the velocity of the other ball?
View Full Question and Explanation
Question 4
Difficulty - Intermediate
Solve Type - Mathematical
A baseball, mass 0.5 kg, is traveling to the right at 32.2 m/s when it is hit by a bat and travels the opposite direction at 72.2 m/s. The bat hits the ball with a force of 1,222 N. What is the ball’s change in momentum and how long was the ball in contact with the bat?
View Full Question and Explanation
Question 5
Difficulty - Advanced
Solve Type - Mathematical
A 4 kg mass is traveling at 10 m/s to the right when it collides elastically with a stationary 7 kg mass. The 7 kg mass then travels at 2 m/s at an angle of 22° below the horizontal. What is the velocity of the 4 kg mass?
View Full Question and Explanation
Question 6
Difficulty - Advanced
Solve Type - Mathematical
A 3800 kg open railroad car coasts along with a constant speed of 8.60 m/s along a level track. Snow begins to fall vertically and fills the car at rate of 3.50 kg/min. Ignoring friction with the tracks, what is the speed of the car after 90 min?
View Full Question and Explanation
Question 7
Difficulty - Advanced
Solve Type - Mathematical
A 2,000 kg car collides with a stationary 1,000 kg car. Afterwards, they slide 6 m before coming to a stop. The coefficient of friction between the tires and the road is 0.7. Find the initial velocity of the 2,000 kg car before the collision?
View Full Question and Explanation
Question 8
Difficulty - Advanced
Solve Type - Conceptual
Consider the following cases of inelastic collisions. Case (1) - A car moving at 75 mph collides with another car of equal mass moving at 75 mph in the opposite direction and comes to a stop. Case (2) A car moving at 75 mph hits a stationary steel wall and rolls backs. The collision time is the same for both cases. In which of these cases would result in the greatest impact force?
  1. Case 1

  2. Case 2

  3. Equal force in both case 1 and case 2

  4. More information is needed to calculate the force

  5. None of these choices.

View Full Question and Explanation
Question 9
Difficulty - Intermediate
Solve Type - Conceptual
A bowling ball moving with speed v collides head-on with a stationary tennis ball. The collision is elastic and there is no friction. The bowling ball barely slows down. What is the speed of the tennis ball after the collision?
  1. nearly v

  2. nearly 2v

  3. nearly 3v

  4. infinite

  5. zero

View Full Question and Explanation
Picture of Jason Kuma
Jason Kuma

Writer | Coach | Builder | Fremont, CA


Units in AP Physics 1

Unit 1 – Linear Kinematics

Unit 2 – Linear Forces

Unit 3 – Circular Motion

Unit 4 – Energy 

Unit 5 – Momentum 

Unit 6 – Torque 

Unit 7 – Oscillations 

Unit 8 – Fluids

Reading Key


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Made By
\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)


Deca- or Deka-


















  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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