# Momentum (12) (ELCA)

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## Background InformationEdit

Kinematic Equations describe the motion of a ball before and after it is kicked. Newton's laws explain why the motion of the ball changes. Momentum explains how the force and the duration of the collision between the ball and the person who kicked it affect the motion of the ball. Momentum changes when a net force is applied. Momentum changes when force changes, quick changes relate to large forces.

## MomentumEdit

• Momentum (p) – the mass of the object multiple by its velocity.

Momentum is so fundamental that it was viewed by Newton as the quantity of motion. It was referred to as progress, thus the p.
$p = mv$

• $p$ – Momentum ($kg\frac{m}{s}$)
• $m$ – Mass (kg)
• $v$ – Velocity ($\frac{m}{s}$)

$F \Delta t =\Delta p = m (v_f - v_i)$

• $F$ – Force (N)
• $\Delta t$ – Chnage in Time (s)
• $\Delta p$ – Change in Momentum (kg$\frac{m}{s}[itex]) *[itex]m$ – mass (kg)
• $v_f$ – Final Velocity ($\frac{m}{s}$)
• $v_i$ – Initial Velocity ($\frac{m}{s}$)

## Momentum and Newton's Second Law of MotionEdit

Newton's Second Law ($F = ma$) tells us that acceleration is high than force is high. Rapid changes in momentum imply large acceleration, and result in large acceleration and therefore, large force.

Application of this law is in a car crash. In a car crash, momentum rapidly drops to zero as the car body crumples. Car bodies crumple slowly, seat belts stretches slightly, and air bags are deployed in order to slow are person down.

Acceleration is $\frac{\Delta v}{\Delta t}$

• $\Delta v$ – Change in velocity ($\frac {m}{s}$)
• $\Delta t$ – Change in time (s)

So, $F = m \frac{\Delta v}{\Delta t}$

• $m$ – Mass (kg)
• $\Delta v$ – Change in velocity ($\frac {m}{s}$)
• $\Delta t$ – Change in time (s)

Therefore, $F = \frac {\Delta p}{\Delta t}$

• $\Delta p$ – Change in momentum ($kg\frac {m}{s}$)
• $\Delta t$ – Change in time (s)

Newton's second law was originally written as $F = \frac {\Delta p}{\Delta t}$. Force equals the rate of change of momentum.

## Impulse-Momentum TheoremEdit

Change in momentum equals force multiplied by a time interval.

$F \Delta t = \Delta p$

• $F$ – Force (N)
• $\Delta t$ – Change in Time (s)
• $\Delta p$ – Momentum ($kg \frac {m}{s}$)

$F \Delta t = \Delta p = mv_f - mv_i$

• $F$ – Force (N)
• $\Delta t$ – Change in Time (s)
• $\Delta p$ – Momentum ($kg \frac {m}{s}$)
• $m$ – Mass (kg)
• $v_f$ – Velocity ($\frac{m}{s}$)
• $v_i$ – Velocity ($\frac{m}{s}$)

$F \Delta t$ is called the impulse of the force $F$ for the time interval $\Delta t$. Extending the time interval over which a constant force is applied allows a smaller force to cause a greater change in momentum than if the same force were applied for a short time period.

### Real Life ApplicationEdit

Highway safety engineers use the impulse-momentum theorem to determine safe following distance for cars and trucks. A fully loaded dump truck has twice the mass as an empty dump truck. If the two dump trucks are traveling at the same speed, the loaded truck has twice the momentum and will take twice as long to stop as the unloaded dump truck.

### Practice ProblemEdit

A 2250 kg car traveling to the west slows down uniformly from 20.0 $\frac{m}{s}$ to 5.00 $\frac{m}{s}$. How long does it take the car to decelerate if the force on the car is 8450 N to the east? How far does the car travel during the deceleration.

• m = 2250 kg
• $v_i$ 20.0 $\frac{m}{s}$ west = -20.0 $\frac{m}{s}$
• $v_f$ 5.00 $\frac{m}{s}$ west = -5.00 $\frac{m}{s}$
• F = 8450 N east
• t = ?
• $\Delta t = \frac {2250 kg (-5 \frac{m}{s} - -20 \frac{m}{s} } {8450 kg \frac{m} {s^2} } = 4 s$
• $\Delta x = \frac {1}{2}(-20 \frac {m}{s} + -5 \frac{m}{s}) 4s = -50 m$

Starting from rest, an 1,800 kg rocket takes off, ejecting 100 kg of fuel per second out of its nozzle at a speed of 2,500 $\frac{m}{s}$. Calculating the force on the rocket from the change in momentum of the fuel.

• $M_r$ = 1,800 kg
• $M_f$ = 100 kg
• $V_\text{ff}$ = 2500 $\frac{m}{s}$
• $F = \frac{\Delta p}{\Delta t} = \frac{250,000 kg \frac{m}{s}}{1s} = -250,000 N$

## Conservation of MomentumEdit

The total momentum of two or more objects interacting with each other remains constant. Total initial momentum is equal total final momentum.
$m_\text{a}v_\text{ai} + m_\text{b}v_\text{bi} = m_\text{a}v_\text{af} + m_\text{b}v_\text{bf}$

• $m_\text{a}$ – mass of object a
• $v_\text{ai}$ – initial speed of object a
• $v_\text{af}$ – final speed of object a
• $m_\text{b}$ – mass of object b
• $v_\text{bi}$ – initial speed of object b
• $v_\text{bf}$ – final speed of object b

Momentum is conserved when the two balls collide or when objects are pushed away from each other.

### Practice ProblemEdit

A 76 kg person, initially at rest in a stationary 45 kg boat, steps out of the boat and onto the dock. It the person moves out of the boat with a velocity of 2.5 $\frac{m}{s}$ to the right, what is the final velocity of the boat.

$m_\text{a}v_\text{ai} + m_\text{b}v_\text{bi} = m_\text{a}v_\text{af} + m_\text{b}v_\text{bf}$

$(76 kg) (0 \frac{m}{s}) + (45 kg) (0 \frac{m}{s}) = (76 kg) (2.5 \frac{m}{s}) + (45 kg) v_\text{bf}$

$0 = 190 kg\frac{m}{s} + 45 kg (v_\text{bf})$

$-190 kg\frac{m}{s} = 45 kg (v_\text {bf})$

$v_\text{bf} = \frac{190 kg\frac{m}{s}}{45 kg}$

$v_\text{bf} = -4.2 \frac{m}{s}$ or $4.2 \frac{m}{s}$ left

## Momentum Change and CollisionsEdit

Compare these scenarios:

• A rubber ball is bounced on the gym floor and bounces upward.
• A clay ball is bounced on the gym floor and hits with a thud.

There are two types of collision:

1. Two objects collide and stick together so that they travel after impact.
2. Two objects collide and move apart with two different velocities.

The total momentum remains constant (see Conservation of Momentum) in any type of collision. The total kinetic energy is generally not conserved in a collision because some kinetic energy is converted to internal energy when the objects deformed.

• Perfectly Inelastic Collisions – two objects collide and move together as one mass.
• Example: arrow hitting target, meteorite colliding and embedding in the Earth's surface.

Since the objects become essentially one object after the collision, the final mass is equal to the combined mass of the two objects, and they move with the same velocity after colliding.
$m_1 v_\text{1i} + m_2 v_\text{2i} = (m_1 + m_2)v_f$

• $m_1$ – Mass of the first object (kg)
• $v_\text{1i}$ – Intial Velocity of the First Object (m/s)
• $m_2$ – Mass of the second object (kg)
• $v_\text{2i}$ – Velocity of the second object (m/s)
• $v_f$ – final velocity (m/s)

### Practice ProblemsEdit

A 1850 kg luxury sedan stopped at a traffic light is struck from the rear by a compact car with a mass of 975 kg. The two cars become entangled as a result of the collision. If the compact car was moving at a velocity of 22.0 m/s to thye north before the collision, what if the velocity of the entangled mass after the collision?

• $m_s$ – 1850 kg
• $v_\text{si}$ – 0 m/s
• $m_c$ – 975 kg
• $v_\text{ci}$ – 22 m/s
• $v_f$ – ?

$m_1 v_\text{1i} + m_2 v_\text{2i} = (m_1 + m_2)v_f$

$0 kg\frac{m}{s} + (975 kg)(22 \frac{m}{s}) = (1850 kg + 975 kg) v_f[itex]

[itex]21450 kg\frac{m}{s} = (2825 kg) v_f$

$v_f = \frac{21450 kg \frac{m}{s}}{2825 kg} = 7.59 \frac{m}{s}$ North

## Kinetic Energy and MomentumEdit

The total kinetic energy does not remain constant when the object collide and stick together. Energy is transformed to sound energy and internal energy as the objects deform during the collision.

### Practice ProblemsEdit

Two clay balls collide in a perfect inelastic collision. The mass of the first ball is .500 kg and has an initial velocity of 4.00 m/s to the right. The mass of the second ball is .250 kg and and initial velocity of 3.00 m/s to the left. What is the final velocity of the composite ball of clay after the collision.

• $m_\text{1}$ – .500 kg
• $v_\text{1i}$ – 4.00 $\frac{m}{s}$
• $m_\text{2}$ – .250 kg
• $v_\text{2i}$ – -3.00 $\frac{m}{s}$
• $v_\text{f}$ – ?

$v_f=\frac{(.500 kg)(4 \frac{m}{s} ) + (.250 kg)( -3.00 \frac{m}{s})} {.750 kg} = 1.67 \frac{m}{s}$

How much does the kinetic energy decrease?

• $m_\text{1}$ – .500 kg
• $v_\text{1i}$ – 4.00 $\frac{m}{s}$
• $m_\text{2}$ – .250 kg
• $v_\text{2i}$ – -3.00 $\frac{m}{s}$
• $v_\text{f}$ – 1.67 $\frac{m}{s}$

$KE_i = \frac{1}{2}m_\text{b1}v_\text{1i}^2 + \frac{1}{2}m_\text{b2}v_\text{2i}^2$

$KE_i = \frac{1}{2}(.500 kg) (4.00 \frac{m}{s})^2 + \frac{1}{2}(.250 kg) (-3.00\frac{m}{s})^2 = 5.13 J$

$KE_f = \frac{1}{2}(m_\text{bi} + m_\text{b2})v_f^2$

$KE_f = \frac{1}{2}(.500 kg + .250)1.67^2 = 1.05$

$\Delta KE = KE_f - KE_f$

$\Delta KE = 1.05 J - 5.13 J = -4.08 J$

## Elastic CollisionsEdit

Elastic Collision occur when the objects remain separate after the collision, like a soccer player kicking a soccer ball. The two objects collide and return to their original shapes with no change in total kinetic energy. In an elastic collision both the total momentum and the total kinetic energy will remain constant. Momentum and Kinetic Energy remain constant in an elastic collision, therefore one can use either the Conservation of Kinetic Energy or the Conservation of Momentum to solve for equations involving elastic collisions.
$m_1v_\text{1i}+m_2v_\text{2i}=m_2v_\text{1f}+m_2v_\text{2f}$

$\frac{1}{2}m_1v_\text{1i}^2 + \frac{1}{2}m_2v_\text{2i}^2 = \frac{1}{2}m_1v_\text{1f}^2 + \frac{1}{2}m_2v_\text{2f}^2$

### Practice ProblemEdit

A .015 kg marble moving to the right at .225 m/s makes an elastic collision with a .03 kg shooter marble moving .18 m/s left. After the collision, the smaller marble moves .315 m/s to the left. Assuming neither marble rotates and is moving on a frictionless surface, what is the final velocity of the shooter marble.

• $m_m$ – .015 kg
• $v_\text{mi}$ – .225 m/s
• Failed to parse (lexing error): m_s[itex] – .030 kg *[itex]v_\text{si}
– - .180 m/s

• $v_\text{mf}$ – - .315 m/s
• $v_\text{sf}$ – ?

$(.015 kg)(.225 \frac{m}{s}) + (.030 kg) (.180 \frac{m}{s}) = (.015 kg) (- .315 \frac{m}{s}) + (.030 kg) (v_\text{sf} \frac{m}{s})$

$v_\text{sf} = \frac{.0027 kg \frac{m}{s}}{.03 kg} = .09 \frac{m}{s}$

## Inelastic Collisions vs. elastic collisionsEdit

### Which ball has the greater force on the floor?Edit

Assume both balls have a mass of 1 kg and both balls are moving at 2 m/s when they hit the floor.
A rubber ball goes from - 2 kg$\frac{m}{s}$ to + 2 kg$\frac{m}{s}$, a change of +4 kg$\frac{m}{s}$.
A clay ball goes from - 2 kg$\frac{m}{s}$ to 0 kg$\frac{m}{s}$, a change of +2 kg$\frac{m}{s}$.
Therefore, the rubber ball has a greater change of momentum.

### Bounces and MomentumEdit

Change in momentum is always greater when objects bounce compared to no bounce. A change in momentum of + 4 kg $\frac{m}{s}$ only tells you the product of force and time, not force or time individually. The product of a force and the time the force acts is called the impulse.

## ImpulseEdit

Rearrange the momentum form of the $\text{2}^\text{nd}$ law to get $F \Delta t = \Delta p$. Impulse ($F \Delta t$) is measured in Ns. Change in momentum is still kg$\frac{m}{s}$.

## Real Life CollisionsEdit

Most collisions in the real work are neither elastic nor perfectly inelastic. Even nearly elastic collisions (between billiards balls or between a player's foot and a ball) result in some decrease in kinetic energy. In real life, some kinetic energy is transformed into sound, heat (from friction), and internal elastic energy (deformation of the ball).

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