The pushing interaction between car tires and the road allows a car to change its speed.
rocket flight was considered impossible less than 100 years ago.
The article object's motion is described in Section 1.7.
He was ridiculed by the press for suggesting a rocket flight to the Moon.
The mass is the same in the closed flask as it is in the open one.
We need quantitative information about the forces that objects exert on the steel wool to use this law.
The pieces are flying apart because of the forces that are exerted on them.
When the forces are not known, the block balances the steel wool nomena.
There is a steel wool block and a flask.
The physical quantity of mass is what we begin our investigation with.
The greater the object's mass, the less it accelerated due to an unbalanced exter nal force.
The steel wool seems to change after being burned.
The steel wool is cold.
The mass of a seedling increases as the plant grows.
There is a closed flask.
The increased mass still balances.
What happens to the burning log is explained by a system perspective.
Air is needed for burning.
The mass is when the steel wool is burned.
The mass in the open flask increases when we burn the steel wool.
The steel wool increases in strength.
The choice of the system was very impor tant to him.
The mass of an isolated system is the mass of all objects in the system.
The mass might change if the system is not isolated.
The change is always equal to the amount of mass leaving the environment.
The change of mass is described by the above equation.
If there is no flow of mass in or out of the system, the mass changes in a predictable way.
Mass does not disappear without a trace.
A bar chart can be used to represent this process.
The bar on the left shows the initial mass of the system, the central bar shows the mass added or taken away, and the bar on the right shows the mass in the final situation.
A mass bar chart shows a constant quantity.
The law of constancy of mass in an iso lated system does not apply in all cases.
Mass is not always constant in an isolated system, but it is a new quantity that includes mass as a component.
The mass of wood decreases when you burn a log in a fire pit.
Mass is an example of a conserved quantity.
There is a transfer of motion from your foot to the ball when you kick a stationary bal.
A similar transfer occurs when you knock bowling pins down.
Let's conduct some experiments to find out.
Both carts will be included in the system for these experiments.
A collision is when two objects come into contact with each other.
The system is isolated because the internal forces that the carts exert on each other are either balanced or negligible.
All of the velocities are with respect to the track in the system of two carts.
1.0 m/s right at 1.0 m/s collides with cart B, which is stationary.
B stops and cart B moves quickly.
0.20 kg1+1.0 m>s2 + 0.20 kg102 are the same as before and after the collision.
A at 1.2 m/s and a cart at 0.4 m/s.
Before and after the collision, the component of velocity is the same.
A piece of clay attached to the front moves at 1.0 m/s.
A move left at 1.0 m/s.
0.20 kg1+1.0 m>s2 + 0.20 kg1-1.0 m>s2 is the same as before and after the collision.
Before and after the carts collided, the same thing happened.
A is moving left.
The two carts should move right at a speed of about 0.33 m/s after the collision.
The outcome was the same as our prediction.
There is a quantity with components.
There are three important points to note.
It is important to consider the direction in which the objects are moving before and after a collision.
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Different observers will measure different momenta for the same object because the veloc ity depends on the choice of the reference frame.
The mo mentum of a car with respect to you is zero.
An observer on the ground can see the car moving away from him.
The two carts were chosen for our investigation.
The system we chose was isolated because the sum of the external forces was zero.
It appears that the total momentum of an isolated system is constant, based on the results of Table 5.1 and Table 5.2.
It's because momentum is a quantitive quantity.
For a system with more than two objects, we simply include a term on each side of the equation for each object.
The idea that the momentum of an isolated system is constant in another situation is being tested.
Jen and David push each other off the rollerblades.
Rollerbladers push fast.
How far is it from the right?
The motion traveled 3.0 m with respect to the floor.
The two rollerbladers are the system.
The two rollerbladers are resting.
Jen and David moved to the left and right after pushing off.
We assume that the distance he will travel during erted on the rol erbladers does not affect their motion because we model each person as a stancy to calculate David's velocity component and point-like object.
There are no external forces on the system.
The forces that the rol erbladers exert are internal and should not affect predict that David will travel 2.0 m in the positive direc of the system.
When walking and jogging, estimate the magnitude of your momen tum.
The direction we choose is positive.
The initial velocity of each person is zero, the above 1 to 2 m/s.
160 m>s2 90 m>s.
We didn't need to know anything about the forces involved.
The forces they exert on each other were not constant.
The equations we have used up to this point assumed constant forces and acceleration.
We owe it to ourselves to use the idea of momentum constancy to analyze a situation.
We've investigated situations involving isolated systems so far.
In the next section, we will look at momentum in nonisolated systems.
Newton's laws can be used to derive an expres sion.
You exert force on a bowling ball when you push it.
There are two expressions for an object's acceleration.
The left side of the equation shows the change in momentum of the object.
The change depends on the product of the net external force and the time interval during which the forces are exerted on the object.
Newton's second law is written in a different form and involves the physical quantity momentum.
A large force for a short time interval can change the momentum of an object by the same amount as a small force for a long time interval.
When you kick a football or hit a baseball with a bat, your foot or bat can make an impulse on the bal.
The direction of the force is pointed in the direction of impulse.
It is difficult to measure the net average force during a time interval.
We can determine the net force on the right side of the equation.
A powerful tool for analyzing interactions between objects is provided by the combination of impulse and momentum change.
We can now write.
A few points are important.
Vec tor equations are difficult to understand.
We will use the component forms of Eq.
The time interval in the equation is very long.
The longer that object 2 exerts the force on object 1, the greater the momen- tion with the apple is.
This explains why a fast moving object might have less of a small impulse than a slow moving object.
A fast moving bul et passing through a partial y closed wooden door might not open the door because it will just make a hole in the door, whereas your little finger, moving much slower than the bullet, could open the door.
The time interval during which the bul let moves at high speed and exerts a large force on the door is very small.
It exerts a relatively small impulse on the door.
The apple's support was not knocked off by the bul et's impulse.
The average force is used if the magnitude of the force changes during the time interval.
The large-mass object would experience a greater change in speed than the smal-mass object.
A person is travelling in a car that is moving to stop him.
16 m/s with respect to the ground when the car hits a bar rier.
Determine the sketch of the process after drawing an initial-Final by an air bag.
Since we are investigating a force being exerted on him, we chose the person as the system.
A crash test dummy is in a car.
The force that the hard surface exerts on the dummy would be about 50,000 N.
The why air bags save lives can be solved.
Let's apply the impulse-momentum equation.
We first look at each cart as a separate system and then look at them together as a single system.
2 on 1 1 on 2 carts do not affect their motion.
This analysis is repeated with cart 2.
The same equation we arrived at in Section 5.2 was used to understand the constant momentum of an isolated sys tem.
The same conclusions have been reached using only our knowledge ofNewton's laws, momentum, and impulse.
An apple is in a tree.
The force responsible should be specified.
There is a system in which the momentum is constant.
We can summarize what we have learned.
The net external impulse on the system is equal to the change in momentum.
The momentum of the system is constant if the net impulse is zero.
We can use the generalized impulse-momentum principle to treat momentum as a conserved quantity.
Chapter 5 is useful in two ways.
We can start from a single principle, regardless of the situation, if we choose to analyze a situation using the ideas of impulse and momentum.
The equations remind us that we need to consider all the interactions between the environment and the system that might cause a change in the system's momentum.
The impulse-momentum process can be described using Eqs.
The equations help us see that we can use a bar chart to represent the changes of a system's mass.
A bar chart made from qualitative impulse-momentum.
Pick the initial and final states, and then choose a system.
We represent the process in an initial- final sketch before constructing the bar chart.
We use the sketch to help us with the impulse-momentum bar chart.
In the final state shown, the carts are stuck together and moving in a positive direction.
They have the same final momentum because they have the same mass.
The shading reminds us that impulse does not reside in the system, it is the influence of the external objects on the system's momentum.
The shaded impulse bar should equal the sum of the heights of the bars on the left and right.
The "conservation of bar heights" is a reflection of momentum.
The bar chart can be used to apply the generalized impulse-momentum equation.
The sign of the term depends on the orientation of the bar.
Car 2 exerts an impulse on car 1 during the collision.
The total height of the initial momentum bar on the left side of the chart and the height of the impulse bar add up to the total height of the final momentum bar on the right side.
The initial and final momentum are positive.
The reference frame is the object of reference and the coordinate system.
The direction of the bars on the bar chart should match the impulse based on the coordinate system.
The final state is after you have two identical mass and size that behave like the board.
The happy bal is very different.
The sad bal does not drop when you do.
The bal bounces back almost the same height as it was dropped.
Happy ball for both balls just before place a wood board on its end and then hit the board below the support for each hitting the board string.
If you want to release the bals one at a time, you have to pul each bal back to its normal height.
The sad ball has the best chance of stopping the board when it hits the board.
The bal is the system.
An expression for change is now possible.
The third law states that each ball exerts an impulse on the board that it hits.
There is a bar chart for each ball-board collision.
The board exerts more force on the happy bal than on the sad bal because it causes the happy bal's momentum to change by more than the sad ball.
The happy bal exerts twice as much force on the board as the sad bal.
The happy bal has a better chance of tipping the board.
Is it riskier for a football player to go to the sad ball.
The impulse-momentum Eq.
is better than any col ision.
When an object bounces back after a col ision, we know that a larger magnitude force is exerted on it than if the object had stopped.
Bulletproof vests for law enforcement agents are designed so that they don't bounce off of it.
Bar charts and initial sketches are useful tools to help analyze problems using the impulse-momentum principle.
Let's look at how these tools work together.
A bul et traveling horizontal y at 250 m is embedded in a block of wood on a table.
Determine the speed of the bul et and wood block together.
The left side of the sketch shows the bul et traveling in the appropriate coordinate axes.
It joins the wood with respect to the ground.
The object of reference is Earth.
Choose a system based on the final state is immediately after the initial state.
Determine if there is any external force exerted on the system.
The very force diagram can be used to determine the short col ision time interval.
The process is represented by the bar chart.
We don't draw a force diagram because the system is isolated.
The sign in front of the equation is the orientation of the bar de term.
The plus or minus signs of the compo nents are based on the chosen coordinate system.
The magnitude of the answer seems reasonable, given how fast the respect to sign, unit, and magnitude is.
It makes sense to make sure it applies for limiting.
The units are correct.
If the mass of the bullet is zero, the very large mass, we can use a limiting case.
A bullet is fired into a block of wood on a table.
To determine the initial speed of the bullet before hitting the block, we could have worked example 5.4 backwards.
It is difficult to measure the bul et's speed since it travels so fast.
Variations of this method are used to decide if golf balls conform to the rules.
The balls are hit by a mechanical launching impulse and the bal s are hit by another object.
The balls' speeds are determined by the speed of the object they are in.
We can estimate the stopping time interval by estimating the stopping distance of the system object.
The car's stopping distance is the distance from the beginning of the impact to the end.
A rough estimate of the stopping distance of a meteorite is provided by the depth of its hole.
We need the stopping time interval associated with the collision, not the stopping distance, to use the impulse momentum principle.
We can estimate the stopping time interval using a known stopping distance.
The acceleration of the object is constant.
There is an equation that can be applied to horizontal or vertical stopping.
The record for the highest movie stunt fall without a parachute is 71 m, held by A. J. Bakunas.
His fall was stopped by a large air cushion.
Estimate the force that the cushion exerts on his body.
The part that causes an impulse is what we focus on.
Below is a sketch of the situation.
The axis is pointing up.
The motion is with respect to the Earth.
The fig final state gives a lot of information about the process.
It is important to pay attention to the signs of the quantities.
The force we are trying to estimate is the average normal force that the cushion exerts.
We draw a force diagram top equation and get it right.
The net force that was put on him points upward, in order to stop Bakunas.
The stopping time interval is used to make automobiles safer for passengers.
It is easy to make sign mistakes.
If the stopping time interval were shorter, the force exerted by the air cushion would be even greater.
If stunt divers practice landing in the last jury, the stopping example would be stopped in 1.0 m instead of 4.0 m.
The stopping time interval is 0.056 s, and the plied to developing air bags and collapsible frames for average stopping force is 50,000 N.
In the previous example, we used a strategy to analyze skull injuries that might lead to concussions.
The human skul can break if the force on it per unit area is 1.7.
The surface area of the skul is less than 1 m2, so we will use square centimeters.