A. elastic collisions.
B. inelastic collisions.
C. perfectly elastic collisions.
D. perfectly inelastic collisions.

Answer) B.

Q2) A helium atom collides with another helium atom in an elastic collision. Which of the following is true?

A. Both momentum and kinetic energy are conserved.
B. Momentum is conserved but kinetic energy is not conserved.
C. Kinetic energy is conserved but momentum is not conserved.
D. Neither momentum nor kinetic energy is conserved.

Q) A softball player leaves the batter's box, overruns first base by 3.0 meters, and then returns to first base. Compared to the total distance traveled by the player, the magnitude of the player's total displacement from the batter's box is

A. Smaller B. Larger C. The same

Answer) A.

Q) A girl leaves a history classroom and walks 10. meters north to a drinking fountain. Then she turns and walks 30. meters south to an art classroom. What is the girl's total displacement from the history classroom to the art classroom?

A. 20. m south B. 20. m north C. 40. m north D. 40. m south

Answer) A.

Distance and Displacement difference Simulation(Virtual Experiment)

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Distance, Position, and Displacement

Motion is a change in the location of an object, as measured by an observer. Distance, in physics terms, means the total length of the path travelled by an object in motion. The SI metric base unit for distance is the metre (m). To help you understand the terms that describe motion, imagine that you are at your home in Figure 1.

Figure 1 Distance and direction along a straight line

You are at the location marked “0 m.” If you walk directly from home to your school in a straight line, you will travel a distance of 500 m.

If you walk from your school to the library and then return home, you will travel an additional distance of 700 m + 1200 m = 1900 m.

If your friend wants to know how to get to the library from your home, telling him to walk for 1200 m is not very helpful. You also need to tell your friend which direction to go. Direction is the line an object moves along from a particular starting point, expressed in degrees on a compass or in terms of the compass points (north, west, east, and south). Directions can also be expressed as up, down, left, right, forward, and backwards. Directions are often expressed in brackets after the distance (or other value). For example, 500 m [E] indicates that the object is 500 m to the east.

Direction is important when describing motion. If the school in Figure 1 is your starting point, the library is in a different direction from your school than your home is. If the library is your starting point, then your school and home are in the same direction.

Scalar and vector Quantities

A scalar quantity is a quantity that has magnitude (size) only. Distance is an example of a scalar quantity. Since direction is so important in describing motion, physicists frequently use terms that include direction in their definitions. A vector is a quantity that has magnitude (size) and also direction. An arrow is placed above the symbol for a variable when it represents a vector quantity.

Position and displacement

Position is the distance and direction of an object from a particular reference point. Position is a vector quantity represented by the symbol $\vec{d}$. Notice the vector arrow above the symbol d. This arrow indicates that position is a vector: it has a direction as well as a magnitude. For example, if home is your reference point, the position of the school in Figure 2 is 500 m [E]. Note that the magnitude of the position is the same as the straight-line distance (500 m) from home to school, but the position also includes the direction (due east [E]). The position of the school from point 0 m can be described by the equation

$\vec{d}_{school}$= 500 m [E]

Now assume that the library is your reference point, or the point 0 m. The position of the school from the reference point (library) can be described by the equation

$\vec{d}_{school}$= 500 m [E]

Once the position of an object has been described, you can describe what happens to the object when it moves from that position. This is displacement—the change in an object’s position. Displacement is represented by the symbol $\vec{\Delta d}$ .

Notice the vector arrow indicating that displacement is a vector quantity. The triangle symbol Δ is the Greek letter delta. Delta is always read as “change in,” so $\vec{\Delta d}$ is read as “change in position.” As with any change, displacement can be calculated by subtracting the initial position vector from the final position vector:

When an object changes its position more than once (experiences two or more displacements), the total displacement $\vec{d}_{T}$ of the object can be calculated by adding the displacements using the following equation:

Q1) Most collisions in the everyday world are
A. elastic collisions.
B. inelastic collisions.
C. perfectly elastic collisions.
D. perfectly inelastic collisions. Answer) B. Q2) A helium atom collides with another helium atom in an elastic collision. Which of the following is true?
A. Both momentum and kinetic energy are conserved.
B. Momentum is conserved but kinetic energy is not conserved.
C. Kinetic energy is conserved but momentum is not conserved.
D. Neither momentum nor kinetic energy is conserved. Answer) A.

Elastic Collisions in One Dimension Simulation(Virtual Experiment)

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In elastic collisions, both kinetic energy and momentum are conserved.

Energy and momentum are always conserved in a collision, no matter what happens. Momentum is easy to deal with because there is only “one form” of momentum, (p=mv), but you do have to remember that momentum is a vector. Energy is tricky because it has many forms, the most troublesome being heat, but also sound and light. If kinetic energy is conserved in a collision, it is called an elastic collision. In an elastic collision, the total kinetic energy is conserved because the objects in question “bounce perfectly” like an ideal elastic. An inelastic collision is one where some of the of the total kinetic energy is transformed into other forms of energy, such as sound and heat.

Any collision in which the shapes of the objects are permanently altered, some kinetic energy is always lost to this deformation, and the collision is not elastic. It is common to refer to a “completely inelastic” collision whenever the two objects remain stuck together, but this does not mean that all the kinetic energy is lost; if the objects are still moving, they will still have some kinetic energy.

Figure 1. Body 1 moves along an x axis before having an elastic collision with body 2, which is initially at rest. Both bodies move along that axis after the collision.

Figure 1 shows two bodies before and after they have a one-dimensional collision, like a head-on collision between pool balls. A projectile body of mass $\ m_{1}$ and initial velocity $\ v_{1i}$ moves toward a target body of mass $\ m_{2}$ that is initially at rest ($\ v_{1i}$ =0). Let’s assume that this two-body system is closed and isolated. Then the net linear momentum of the system is conserved, and from Eq. 9-51 we can write that conservation as

In each of these equations, the subscript i identifies the initial velocities and the subscript f the final velocities of the bodies. If we know the masses of the bodies and if we also know $\ v_{1i}$ , the initial velocity of body 1, the only unknown quantities are $\ v_{1f}$ and $\ v_{2f}$, the final velocities of the two bodies. With two equations at our disposal, we should be able to find these two unknowns.

Note that $\ v_{2f}$ is always positive (the initially stationary target body with mass $\ m_{2}$ always moves forward). From Eq. (5) we see that $\ v_{1f}$ may be of either sign (the projectile body with mass $\ m_{1}$ moves forward if $\ m_{1} > m_{2} $ but rebounds if $\ m_{1} < m_{2} $).

Let us look at a few special situations(Elastic Collisions in One Dimension).

1. Equal masses

If $\ m_{1} = m_{2} $, Eqs. (5) and (6) reduce to

$\ v_{1f} = 0 $ and $\ v_{2f} = v_{1i} $

which we might call a pool player’s result. It predicts that after a head-on collision of bodies with equal masses, body 1 (initially moving) stops dead in its tracks and body 2 (initially at rest) takes off with the initial speed of body 1. In head-on collisions, bodies of equal mass simply exchange velocities. This is true even if body 2 is not initially at rest.

2. A massive target

In Fig. 9-18, a massive target means that $\ m_{2} \gg m_{1} $. For example, we might fire a golf ball at a stationary cannonball. Equations (5) and (6) then reduce to

This tells us that body 1 (the golf ball) simply bounces back along its incoming path, its speed essentially unchanged. Initially stationary body 2 (the cannonball) moves forward at a low speed, because the quantity in parentheses in Eq. 9-69 is much less than unity.All this is what we should expect.

3. A massive projectile

This is the opposite case; that is, $\ m_{1} \gg m_{2} $. This time, we fire a cannonball at a stationary golf ball. Equations (5) and (6) reduce to

Equation (8) tells us that body 1 (the cannonball) simply keeps on going, scarcely slowed by the collision. Body 2 (the golf ball) charges ahead at twice the speed of the cannonball. Why twice the speed? Recall the collision described by Eq. (7), in which the velocity of the incident light body (the golf ball) changed from $\ v$ to $\ - v$, a velocity change of $\ 2 v$. The same change in velocity (but now from zero to $\ 2 v$) occurs in this example also.

Q) You row a boat perpendicular to the shore of a river that flows $\ v_{b}$=3 m/s. The velocity of your boat is $\ v_{w}$=4 m/s relative to the water. What is the velocity of your boat relative to the shore?

A. 1 m/s B. 5 m/s C. 7 m/s D. 15 m/s

Answer) B.

Relative velocity of Raindrops Simulation(Experiment)

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Relative Velocity
All velocity is measured from a reference frame (or point of view). Velocity with respect to a reference frame is called relative velocity. A relative velocity has two subscripts, one for the object, the other for the reference frame.

Relative velocity problems relate the motion of an object in two different reference frames.

Relative Velocity of Raindrop
When we train inside and outside of raindrops, then we looked out the window. We see raindrops fall sideways out there. Why is that? This is caused by the existence of two movements moving in different directions.

1. The train moves horizontally with a speed $\ v_{TG}$.

2. Precipitation that falls perpendicular to the ground at $\ v_{RG}$.

If the two vectors that we put down at one point arrested then we will get the sum of the second vector is as below:

$\ v_{TG}$ = Train velocity with respect to Ground
$\ v_{GT} (=-v_{TG}) $= Ground velocity with respect to Train
$\ v_{RG}$ = Rain velocity with respect to Ground
$\ v_{RT}$ = Rain velocity with respect to Train

From the description above it is clear that we see slanting raindrops outside the train is the result of the resultant / sum of two vectors is the speed of the train velocity vector and velocity vector rainwater mutually perpendicular. The second resultant velocity vector is called the relative velocity of raindrops on us as a passenger train.

Q1) Car A is moving at $\ v_{A} $= 60 km/h to the right with respect to the ground. Car B is moving at $\ v_{B} $= 80 km/h to the left with respect to the ground. What is the velocity of Car A with respect to Car B (the velocity of Car A as measured by the passenger in Car B)?

A. 20 km/h, left B. 20 km/h, right C. 60 km/h, left

D. 140 km/h, left E. 140 km/h, right

Answer) E
The velocity of Car A with respect to Car B ($\ v_{AB}$) is given by: $\ v_{AB} = v_{A} – v_{B} $. $\ v_{AB} $ = 60 km/h [right] – 80 km/h [left]) = 60 km/h –(– 80 km/h) = 140 km/h [right]

Q2) Car A is moving at $\ v_{A}$ = 60 km/h to the right with respect to the ground. Car B is moving at $\ v_{B}$ = 80 km/h to the right with respect to the ground. What is the velocity of Car A with respect to Car B (the velocity of Car A measured by Car B)?

A. 20 km/h, left B. 20 km/h, right C. 60 km/h, left

D. 140 km/h, left E. 140 km/h, right

Answer) A

: The velocity of Car A with respect to Car B ($\ v_{AB}$) is given by: $\ v_{AB} = v_{A} – v_{B} $. $\ v_{AB} $= (60 km/h [right]) – (80 km/h [right]) = (60 km/h – 80 km/h) = -20 km/h [left]

Relative velocity one dimension Simulation(Virtual Experiment)

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Relative velocity

The case of the faster car overtaking your car was easy to solve with a minimum of thought and effort, but you will encounter many situations in which a more systematic method of solving such problems is beneficial. To develop this method, write down all the information that is given and that you want to know in the form of velocities with subscripts appended.

$\ v_{se} $ = +80 km/h, north

(Here the subscript se means the velocity of the slower car with respect to Earth.)

$\ v_{fe} $ = +90 km/h, north

(The subscript fe means the velocity of the fast car with respect to Earth.)

We want to know $\ v_{fs} $, which is the velocity of the fast car with respect to the slower car. To find this, we write an equation for $\ v_{fs} $ in terms of the other velocities, so on the right side of the equation the subscripts start with f and eventually end with s. Also, each velocity subscript starts with the letter that ended the preceding velocity subscript.

$\ v_{fs} = v_{fe} + v_{es} $

The boldface notation indicates that velocity is a vector quantity. This approach to adding and monitoring subscripts is similar to vector addition, in which vector arrows are placed head to tail to find a resultant.

We know that $\ v_{es} = −v_{se} $ because an observer in the slow car perceives Earth as moving south at a velocity of 80 km/h while a stationary observer on the ground (Earth) views the car as moving north at a velocity of 80 km/h.

Thus, this problem can be solved as follows:

$\ v_{fs} = v_{fe} + v_{es} = v_{fe} - v_{se} $

$\ v_{fs} $ = (+90 km/h north) − (+80 km/h north) = +10 km/h, north

When solving relative velocity problems, follow the above technique for writing subscripts. The particular subscripts will vary depending on the problem, but the method for ordering the subscripts does not change. A general form of the relative velocity equation is $\ v_{ac} = v_{ab} + v_{bc} $. This general form may help you remember the technique for writing subscripts.

Example 1.
Illustrates the concept of relative velocity by showing a passenger walking toward the front of a moving train. The people sitting on the train see the passenger walking with a velocity of +2.0 m/s, where the plus sign denotes a direction to the right. Suppose the train is moving with a velocity of +9.0 m/s relative to an observer standing on the ground. Then the ground-based observer would see the passenger moving with a velocity of +11 m/s, due in part to the walking motion and in part to the train’s motion. As an aid in describing relative velocity, let us define the following symbols:

Figure 1 The velocity of the passenger relative to the ground-based observer is vPG. It is the vector sum of the velocity $\ v_{PT}$ of the passenger relative to the train and the velocity $\ v_{TG}$ of the train relative to the ground: $\ v_{PG}= v_{PT}+ v_{TG} $.

$\ v_{PT}$ = velocity of the Passenger relative to the Train = +2.0 m/s

$\ v_{TG}$= velocity of the Train relative to the Ground = +9.0 m/s
$\ v_{PG}$ = velocity of the Passenger relative to the Ground = +11.0 m/s

The vector addition problem this illustrates is

$\ v_{PG}= v_{PT}+ v_{TG} $

If we define the forward direction to be the positive direction,

then, because the vectors we are adding are both in the same direction, we are indeed dealing with that very special case in which the magnitude of the resultant is just the sum of the magnitudes of the vectors we are adding:

$\ v_{PG}= v_{PT}+ v_{TG} $

$\ v_{PG}$ = (2.0 m/s) + (9.0 m/s)= +11.0 m/s
$\ v_{PG}$ = 11.0 m/s in the direction in which the train is traveling

You already know all the concepts you need to know to solve relative velocity problems (you know what velocity is and you know how to do vector addition) so the best we can do here is to provide you with some more worked examples. We’ve just addressed the easiest kind of relative velocity problem, the kind in which all the velocities are in one and the same direction. The second easiest kind is the kind in which the two velocities to be added are in opposite directions.

Example 2.
A train is moving along a ground at a constant 9.0 m/s. a passenger that has a velocity of 2m/s straight
backward, (toward the back of the train). Find the velocity of the passenger, relative to the ground.

Q) Two skaters of different masses prepare to push off against one another. Which one will gain the larger velocity?

A. The more massive one
B. The less massive one
C. They will each have equal but opposite velocities.

Answer) B.
Both must move with momentum values equal in magnitude but opposite in direction: $\ m_{1}v_{1} = m_{2}v_{2} $ the skater with the smaller mass must have the larger velocity

Conservation of momentum recoil velocity Simulation(Virtual Experiment)

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Recoil
It is very important to define a system carefully. The momentum of a baseball changes when the external force of a bat is exerted on it. The baseball, therefore, is not an isolated system. On the other hand, the total momentum of two colliding balls within an isolated system does not change because all forces are between the objects within the system. Can you find the final velocities of the two in-line skaters in Figure 1 Assume that they are skating on a smooth surface with no external forces. They both start at rest, one behind the other.

FIGURE 1 The internal forces exerted by these in-line skaters cannot change the total momentum of the system.

Skater A gives skater B a push. Now both skaters are moving, making this situation similar to that of an explosion. Because the push was an internal force, you can use the law of conservation of momentum to find the skaters’ relative velocities. The total momentum of the system was zero before the push. Therefore, it also must be zero after the push.

The coordinate system was chosen so that the positive direction is to the left. The momenta of the skaters after the push are equal in magnitude but opposite in direction. The backward motion of skater A is an example of recoil. Are the skaters’ velocities equal and opposite? The last equation shown above, for the velocity of skater A, can be rewritten as follows:

$\ v_{A2} = -(\frac{m_{B}}{m_{A}}) v_{B_{2}}$

The velocities depend on the skaters’ relative masses. If skater A has a mass of 30.0 kg and skater B’s mass is 60.0 kg, then the ratio of their velocities will be 60/30 or 2.0. The less massive skater moves at the greater velocity. But, without more information about how hard they pushed, you can’t find the velocity of each skater.

Is momentum conserved when shooting a shotgun?

FIGURE 2. The shotgun recoils with a momentum equal in magnitude to the momentum of the shot.
The explosion of the powder causes the shot to move very rapidly forward. If the gun is free to move, it will recoil backward with a momentum equal in magnitude to the momentum of the shot. Even though the mass of the shot is small, its momentum is large due to its large velocity. The shotgun recoils with a momentum equal in magnitude to the momentum of the shot. The recoil velocity of the shotgun will be smaller than the shot’s velocity because the shotgun has more mass, but it can still be sizeable.

How can you avoid a bruised shoulder?
If the shotgun is held firmly against your shoulder, it doesn’t hurt as much. If you think of the system as just the shotgun and the pellets, then your shoulder applies a strong external force to the system. Since conservation of momentum requires the external force to be zero, the momentum of this system is not conserved. If you think of the system as including yourself with your shoulder against the shotgun, then momentum is conserved because all the forces involved are internal to this system (except possibly friction between your feet and the earth). With your mass added to the system, the recoil velocity is smaller. If you think of the system as including yourself and the earth, then momentum is conserved because all the forces involved are internal to this system. The large mass of the earth means that the change in momentum of the earth would be imperceptible.

How does a rocket accelerate in empty space when there is nothing to push against? FIGURE 3. The momentum of the rocket is equal to the momentum of the gases

The exhaust gases rushing out of the tail of the rocket have both mass and velocity and, therefore, momentum. The momentum gained by the rocket in the forward direction is equal to the momentum of the exhaust gases in the opposite direction. The rocket and the exhaust gases push against each other. Newton’s third law applies.