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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: \[p_i = p_f \]

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 \( 0.450 \) \( \text{kg} \) ice puck, moving east with a speed of \( 3.00 \) \( \text{m/s} \), has a head-on collision with a \( 0.900 \) \( \text{kg} \) 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 (\( P_i = P_f\))

\[(0.450)(3.00) = (0.900)(v_1) + (0.450)(v_2) \]

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):

\[(0.450)(3.00)^2 = (0.900)(v_2)^2 + (0.450)(v_1)^2\]

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: \( v_1 = -1 \, \text{m/s} \) (west), \( v_2 = 2 \, \text{m/s} \) (east)

Impulse

Impulse is a change in momentum. This means that \(P_i \neq P_f\).

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

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 \( 0.0600 \) \( \text{kg} \) tennis ball is traveling at \( 30.0 \) \( \text{m/s} \). After being hit by the opponent’s racket, the ball’s velocity is \( 20.0 \) \( \text{m/s} \) 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 \(0.0400\) \(\text{s}\)

Solution:

  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: \( \Delta p = m\Delta v \rightarrow (0.060)(30 + 20) = 3\,\text{N}\cdot\text{s} \) (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. \( \Delta p = Ft \rightarrow 3 = F(0.040) \rightarrow F = 75\,\text{N} \)

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.

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Question 1
Advanced
Mathematical
A bullet of mass \(0.0500 \, \text{kg}\) traveling at \(50.0 \, \text{m/s}\) is fired horizontally into a wooden block suspended from a long rope. The mass of the wooden block is \(0.300 \, \text{kg}\) 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.
Part (a)3 pts

Calculate velocity of the bullet + wooden block just after impact.

Part (b)3 pts

Calculate the vertical distance \(h\) reached by the bullet and wooden block.

Part (c)3 pts

Which of the following two would have the greater effect on the height of the bullet block system: (i) doubling the bullet’s velocity (ii) doubling the bullet’s mass (iii) Neither. Justify your answer.

Open in UBQ
Question 2
Beginner
Mathematical
A child (\(m = 32 \, \text{kg}\)) in a boat (\(m = 71 \, \text{kg}\)) throws a \(7.1 \, \text{kg}\) package out horizontally with a speed of \(12.2 \, \text{m/s}\). Calculate the velocity of the boat immediately after, assuming it was initially at rest. Ignore water resistance.
Open in UBQ
Question 3
Beginner
Mathematical
A pool cue ball, mass \(0.7 \, \text{kg}\), is traveling at \(2 \, \text{m/s}\) when it collides head-on with another ball, mass \(0.5 \, \text{kg}\), traveling in the opposite direction with a speed of \(1.2 \, \text{m/s}\). After the collision, the cue ball travels in the opposite direction at \(0.3 \, \text{m/s}\). What is the velocity of the other ball?
Open in UBQ
Question 4
Intermediate
Mathematical
A baseball, mass \(0.5 \, \text{kg}\), is traveling to the right at \(32.2 \, \text{m/s}\) when it is hit by a bat and travels the opposite direction at \(72.2 \, \text{m/s}\). The bat hits the ball with a force of \(1,222 \, \text{N}\). What is the ball’s change in momentum and how long was the ball in contact with the bat?
Open in UBQ
Question 5
Advanced
Mathematical
A \(4 \, \text{kg}\) mass is traveling at \(10 \, \text{m/s}\) to the right when it collides elastically with a stationary \(7 \, \text{kg}\) mass. The \(7 \, \text{kg}\) mass then travels at \(2 \, \text{m/s}\) at an angle of \(22^\circ\) below the horizontal. What is the velocity of the \(4 \, \text{kg}\) mass?
Open in UBQ
Question 6
Advanced
Mathematical
A \(3800 \, \text{kg}\) open railroad car coasts along with a constant speed of \(8.60 \, \text{m/s}\) along a level track. Snow begins to fall vertically and fills the car at a rate of \(3.50 \, \text{kg/min}\). Ignoring friction with the tracks, what is the speed of the car after \(90 \, \text{min}\)?
Open in UBQ
Question 7
Advanced
Mathematical
A \(2,000 \, \text{kg}\) car collides with a stationary \(1,000 \, \text{kg}\) car. Afterwards, they slide \(6 \, \text{m}\) before coming to a stop. The coefficient of friction between the tires and the road is \(0.7\). Find the initial velocity of the \(2,000 \, \text{kg}\) car before the collision?
Open in UBQ
Question 8
Advanced
Conceptual
Consider the following cases of inelastic collisions. Case (1) – A car moving at \(75 \, \text{mph}\) collides with another car of equal mass moving at \(75 \, \text{mph}\) in the opposite direction and comes to a stop. Case (2) A car moving at \(75 \, \text{mph}\) hits a stationary steel wall and rolls back. The collision time is the same for both cases. In which of these cases would result in the greatest impact force?
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Question 9
Intermediate
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?

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KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

Metric Prefixes

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]

  3. Perform the multiplication: [katex]\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}[/katex]

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

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

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