## Elastic Collision in One Dimension Experiment

#### Quiz

Q1) Most collisions in the everyday world are

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

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.

#### 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

$\ m_{1} v_{1i} = m_{1} v_{1f} + m_{2} v_{2f}$    (linear momentum). --(1)

If the collision is also elastic, then the total kinetic energy is conserved and we can write that conservation as

$\frac{1}{2} m_{1}v_{1i}^2 = \frac{1}{2} m_{1}v_{1f}^2 + \frac{1}{2} m_{2}v_{2f}^2$  (kinetic energy).  --(2)

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.

To do so, we rewrite Eq. (1) as

$\ m_{1}(v_{1i} - v_{1f}) = m_{2}v_{2f}$   --(3)

and Eq. (2) as*

$\ m_{1}(v_{1i} - v_{1f}) (v_{1i} + v_{1f}) = m_{2}v_{2f}^2$  --(4)

After dividing Eq. (4) by Eq. (3) and doing some more algebra, we obtain

$\ v_{1f} = \frac {m_{1} -m_{2}}{m_{1} + m_{2}} v_{1i}$  --(5)

$\ v_{2f} = \frac {2m_{1}}{m_{1} + m_{2}} v_{1i}$  --(6)

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

$\ v_{1f} \approx v_{1i}$   and $\ v_{2f}\approx (\frac{2m_{1}}{m_{2}})v_{1i}$  --(7)

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

$\ v_{1f} \approx v_{1i}$  and  $\ v_{2f} \approx 2 v_{1i}$  --(8)

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.

## Relative Velocity of Raindrops Experiment

#### Quiz

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

#### 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.

## Relative velocity one dimension Virtual Experiment

#### Quiz

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

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

:  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.

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

$\ v_{PG}$=  -$\left | v_{PT} \right |$+ $\ v_{TG}$

$\ v_{PG}$= (-2.0 m/s) + (9.0 m/s)= +7.0 m/s
$\ v_{PG}$ = 7.0 m/s in the direction in which the train is traveling

## Conservation of momentum recoil velocity Virtual Experiment

#### Quiz

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.

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.

BEFORE              AFTER
(State 1)            (State 2)

$\ p_{A1} + p_{B1} = p_{A2} + p_{B2}$
$\ ~~~~~~~ 0~~~~ ~~~~= p_{A2} + p_{B2}$
or: $\ ~~~~ p_{A2} ~~ = -p_{B2}$
$\ ~~~~m_{A}v_{A2} ~= -m_{B}v_{B2}$

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.

## Perfectly inelastic collision Virtual Experiment

#### Quiz

A box slides with initial velocity 10m/s on a frictionless surface and collides inelastically with an identical box.  The boxes stick together after the collision.  What is the final velocity?
A.  10 m/s      B.  20 m/s      C.  0 m/s      D.  15 m/s      E.  5 m/s

The initial momentum is:
$\ Mv_{i} = (10) M$
The final momentum must be the same!!
The final momentum is:
$\ M_{tot} v_{f} = (2M) v_{f} = (2M) (5)$

#### Perfectly inelastic collision Simulation(Virtual Experiment)

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Do the math!
1) Consider a 2-kg fish that swims toward and swallows a 1-kg fish that is swims at –10 m/s. If the larger fish swims at 10 m/s, what is its velocity immediately after lunch?

Momentum is conserved from the instant before lunch until the instant after (in so brief an interval, water resistance does not have time to change the momentum), so we can write

$\ net~ momentum _{before~ lunch} = net~ momentum _{after ~lunch}$
$\ (net ~mv) _{before} = (net~ mv) _{after}$
$\ (2 kg)(10 m/s) + (1 kg)(-10 m/s) = (2 kg + 1 kg)(v_{after})$
$\ 10 kgm/s=(3 kg)(v_{after})$
$\ v_{after}=\frac{10}{3} m/s$

#### Let's study about the following content

Conservation of Momentum
From Newton’s second law you know that to accelerate an object, a net force must be applied to it. This chapter says much the same thing, but in different language. If you wish to change the momentum of an object, exert an impulse on it.
In either case, the force or impulse must be exerted on the object by something outside the object. Internal forces won’t work. For example, the molecular forces within a basketball have no effect on the momentum of the basketball, just as a push against the dashboard of a car you’re sitting in does not affect the momentum of the car. Molecular forces within the basketball and a push on the dashboard are internal forces. They come in balanced pairs that cancel within the object. To change the momentum of the basketball or the car, an outside push or pull is required. If no outside force is present, no change in momentum is possible.

Momentum, like the quantities velocity and force, has both direction and magnitude. It is a vector quantity. Like velocity and force, momentum can be canceled. So, although the cannonball in the preceding example gains momentum when fired and the recoiling cannon gains momentum in the opposite direction, the cannon–cannonball system gains none. The momenta (plural form of momentum) of the cannonball and the cannon are equal in magnitude and opposite in direction. Therefore, these momenta cancel each other out for the system as a whole. No external force acted on the system before or during firing. Since no net force acts on the system, there is no net impulse on the system and there is no net change in the momentum.
In every case, the momentum of a system cannot change unless it is acted on by external forces. A system will have the same momentum before some internal interaction as it has after the interaction occurs.

When momentum, or any quantity in physics, does not change, we say it is conserved. The law of conservation of momentum describes the momentum of a system. The law of conservation of momentum states that, in the absence of an external force, the momentum of a system remains unchanged. If a system undergoes changes wherein all forces are internal as for example in atomic nuclei undergoing radioactive decay, cars colliding, or stars exploding, the net momentum of the system before and after the event is the same.

Collisions
As you go about your day-to-day activities, you probably witness many collisions without really thinking about them. In some collisions, two objects collide and stick together so that they travel together after the impact. An example of this action is a collision between football players during a tackle, as shown in Figure 1.
In an isolated system, the two football players would both move together after the collision with a momentum equal to the sum of their momenta (plural of momentum) before the collision. In other collisions, such as a collision between a tennis racquet and a tennis ball, two objects collide and bounce so that they move away with two different velocities.
The total momentum remains constant in any type of collision. However, the total kinetic energy is generally not conserved in a collision because some kinetic energy is converted to internal energy when the objects deform. In this
section, we will examine different types of collisions and determine whether kinetic energy is conserved in each type.We will primarily explore two extreme types of collisions: elastic and perfectly inelastic collisions.

Perfectly inelastic collisions can be analyzed in terms of momentum
When two objects, such as the two football players, collide and move together as one mass, the collision is called a perfectly inelastic collision. Likewise, if a meteorite collides head on with Earth, it becomes buried in Earth and the collision is perfectly inelastic.

Figure 1. When one football player tackles another, they both continue to fall together. This is one familiar example of a perfectly inelastic collision.

Perfectly inelastic collisions are easy to analyze in terms of momentum because the objects become essentially one object after the collision. The final mass is equal to the combined masses of the colliding objects. The combination moves with a predictable velocity after the collision.
Consider two cars of masses $\ m_{1}$ and $\ m_{2}$ moving with initial velocities of $\ v_{1,i}$ and $\ v_{2,i}$ along a straight line, as shown in Figure 11. The two cars stick together and move with some common velocity, $\ v_{f}$ , along the same line of motion after the collision. The total momentum of the two cars before the collision is equal to the total momentum of the two cars after the collision.

Figure 2. The total momentum of the two cars before the collision (a) is the same as the total momentum of the two cars after the inelastic collision (b).

 PERFECTLY INELASTIC COLLISION $\ m_{1}v_{1,i} + m_{2}v_{2,i} = (m_{1} + m_{2}) v_{f}$

This simplified version of the equation for conservation of momentum is useful in analyzing perfectly inelastic collisions.When using this equation, it is important to pay attention to signs that indicate direction. In Figure 2, $\ v_{1,i}$ has a positive value ($\ m_{1}$ moving to the right), while $\ v_{2,i}$ has a negative value ($\ m_{2}$ moving to the left).

Kinetic energy is not conserved in inelastic collisions
In an inelastic collision, the total kinetic energy does not remain constant when the objects collide and stick together. Some of the kinetic energy is converted to sound energy and internal energy as the objects deform during the collision.
This phenomenon helps make sense of the special use of the words elastic and inelastic in physics.We normally think of elastic as referring to something that always returns to, or keeps, its original shape. In physics, an elastic material is one in which the work done to deform the material during a collision is equal to the work the material does to return to its original shape. During a collision, some of the work done on an inelastic material is converted to other forms of energy, such as heat and sound.

## Ballistic Pendulum Measurement Virtual Experiment

#### Quiz

Q1) A ballistic pendulum is a device used to measure the speed of a bullet. The wooden pendulum bob in such a system absorbs a bullet of known mass moving in a horizontal trajectory. The bullet’s speed can be calculated using the measured angle through which the pendulum swings. Most of the mechanical energy of the flying bullet captured by the pendulum bob in such a device is converted, almost instantly, into

A. mechanical energy of the swinging pendulum bob
B. internal energy of the pendulum bob and bullet
C. sound

Q2) A wood block rests at rest on a table. A bullet shot into the block stops inside, and the bullet plus block start sliding on the frictionless surface. The horizontal momentum of the bullet plus block remains constant
A. Before the collision         B. During the collision        C. After the collision
D. All of the above               E. Only A and C above

As long as there are no external forces acting on the system

#### Ballistic Pendulum Measurement Of Speed Simulation(Virtual Experiment)

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Ballistic Pendulum Measurement Of Speed

The ballistic pendulum is used to measure bullet speeds. The pendulum is a large wooden block of mass $\ M$ hanging vertically by two cords. A bullet of mass $\ m$, traveling with a horizontal speed $\ v_{i}$ strikes the pendulum and remains embedded in it.

If the collision time (the time required for the bullet to come to rest with respect to the block) is very small compared to the time of swing of the pendulum, the supporting cords remain approximately vertical during the collision.

Therefore, no external horizontal force acts on the system (bullet + pendulum) during collision, and the horizontal component of momentum is conserved. The speed of the system after collision $\ v_{f}$ can be easily determined, so that the original speed of the bullet can be calculated from momentum conservation

Figure 1. Ballistic pendulum

The initial momentum of the system is that of the bullet $\ mv_{i}$, and the momentum of the system just after the collision is

$\ (m + M)v_{f}$

Conservation of horizontal linear momentum gives

$\ mv_{i} = (m + M)v_{f}$

After the collision is over, the pendulum and bullet swing up to a maximum height y, where the kinetic energy left after impact is converted into gravitational potential energy. Then, using the conservation of mechanical energy for this part of the motion, we obtain

$\frac{1}{2}(m+M)v_{f}^2 = (m+M)gy$

Solving these two equations for $\ v_{i}$, we obtain

$\ v_{i} = \frac{m+M}{m} \sqrt{2gy}$

Hence, we can find the initial sped of the bullet by measuring $\ m$, $\ M$, and $\ y$.

If the maximum angle the pendulum swings through is θ and the length of pendulum from the center of mass to the point of rotation is $\ L$ , then the height h is given by

$\ y = L(1- cosθ)$

Figure 2. Ballistic pendulum experiment.

It would be difficult to measure the change in height $\ y$ directly. Instead, we will measure the horizontal distance $\ x$ that the pendulum moves, and use this quantity to calculate y. If we consider Figure 2, it is clear the the string length $\ L$ is the hypotenuse of a right triangle with sides $\ (L − y)$ and $\ x$, so:

$\ L^2 = (L-y)^2 + x^2$

If we subtract $\ x^2$ from both sides we obtain

$\ L^2 - x^2 = (L-y)^2$

If we now take the (positive) square root and solve for $\ y$, we find

$\ y = L - \sqrt{L^2 - x^2}$

We will use this expression for $\ y$ in $\ v_{i} = \frac{m+M}{m} \sqrt{2gy}$ above to calculate $\ v_{i}$.

We also want to find the uncertainty in the initial projectile velocity $\ v_{i}$. The measured quantities, $\ x$, $\ L$, $\ m$, and $\ M$ all have some uncertainty, leading to uncertainty in the calculated $\ v_{i}$. The formulas connecting $\ v_{i}$ with these measured quantities would make the error analysis complicated. In addition, another source of uncertainty is in the behavior of the launcher—one sometimes sees variations in $\ v_{i}$ from shot to shot.

## Magnetic force on a current carrying wire virtual experiment

#### Quiz

Q1) The force on a moving charge in the presence of a magnetic field is always

A. Perpendicular to both the charge's velocity and the magnetic field
B. Parallel to the charge's velocity
C. Parallel to the magnetic field
D. None of the above

Q2)  A current is moving from north to south through a long wire that is lying horizontally on a table. What is the direction of the magnetic force if the magnetic field is directed up and out of the table?

A. toward the north
B. toward the east
C. toward the south
D. toward the west

#### Magnetic force on a current carrying wire Simulation(Virtual Experiment)

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Magnetic force on a current carrying wire
If a magnetic field exerts a force on a single charged particle when it moves through a magnetic field, it should be no surprise that magnetic forces are exerted on a current-carrying wire as well (see Fig. 1). Because the current is a collection of many charged particles in motion, the resultant force on the wire is due to the sum of the individual forces on the charged particles. The force on the particles is transmitted to the “bulk” of the wire through collisions with the atoms making up the wire.

FIGURE 1 This apparatus demonstrates the force on a current carrying wire in an external magnetic field. Why does the bar swing away from the magnet after the switch is closed?

Some explanation is in order concerning notation in many of the figures. To indicate the direction of $\ B$, we use the following conventions:

 If $\ B$ is directed into the page, as in Figure 2, we use a series of green crosses, representing the tails of arrows. If $\ B$ is directed out of the page, we use a series of green dots, representing the tips of arrows. If $\ B$ lies in the plane of the page, we use a series of green field lines with arrowheads.

The force on a current-carrying conductor can be demonstrated by hanging a wire between the poles of a magnet, as in Figure 2. In this figure, the magnetic field is directed into the page and covers the region within the shaded area. The wire deflects to the right or left when it carries a current.

FIGURE 2  A segment of a flexible vertical wire partially stretched between the poles of a magnet, with the field (green crosses) directed into the page. (a) When there is no current in the wire, it remains vertical. (b) When the current is upward, the wire deflects to the left. (c) When the current is downward, the wire deflects to the right.

Magnetic Force on a wire resulting from a magnetic field
It is possible to determine the force of magnetism exerted on a current-carrying wire passing through a magnetic field at right angles to the wire. Experiments show that the magnitude of the force, $\ F$, on the wire, is proportional to the strength of the field, $\ B$, the current, $\ I$, in the wire, and the length, $\ L$, of the wire in the magnetic field. The relationship of these four factors is as follows:

 Magnetic Force on a Current-Carrying Wire in a Magnetic Field  $\ F=BIL$ The force on a current-carrying wire in a magnetic field is equal to the product of magnetic field strength, the current, and the length of the wire.

The strength of a magnetic field, $\ B$, is measured in teslas, T. 1T is equivalent to 1 N/A m.
Note that if the wire is not perpendicular to the magnetic field, a factor of $\sin \theta$ is introduced in the above equation, resulting in $\ F=BILsin \theta$ . As the wire becomes parallel to the magnetic field, the angle becomes zero, and the force is reduced to zero. When $\theta$=90°, the equation is again $\ F=BIL$.

Determining the Magnetic force’s direction
Faraday’s description of the force on a current-carrying wire does not completely describe the direction because the force can be upward or downward. The direction of the force on a current carrying wire in a magnetic field can be found by using the third right-hand rule. This technique is illustrated in Figure 3. The magnetic field is represented by the symbol B, and its direction is represented by a series of arrows. To use the third right-hand rule, point the fingers of your right hand in the direction of the magnetic field, and point your thumb in the direction of the conventional (positive) current in the wire.

Figure 3 The third right-hand rule can be used to determine the direction of force when the current and magnetic field are known.

The palm of your hand will be facing in the direction of the force acting on the wire. When drawing a directional arrow that is into or out of the page, direction is indicated with crosses and dots, respectively. Think of the crosses as the tail feathers of the arrow, and the dots as the arrowhead.

## Superposition principle of electric field virtual experiment

#### Quiz

Q1) A has a charge of +2 µC, and object B has a charge of +6 µC. Which statement is true:

A. $\ F_{AB}$= $\ –3F_{BA}$
B. $\ F_{AB}$=$\ –F_{BA}$
C. $\ 3F_{AB}$=$\ –F_{BA}$

By Newton’s third law, the two objects will exert forces having equal magnitudes but opposite directions on each other.

Q2) At the position of the dot, the electric field points

A. Up.
B. Down.
C. Left.
D. Right.
E. The electric field is zero.

The direction of the field is taken to be the direction of the force it would exert on a positive test charge

#### Superposition principle of electric fields Simulation(Virtual Experiment)

When the next simulation is not visible, please refer to the following link.

Force, Charges, and Distance
The electrical force between any two objects obeys a similar inverse-square relationship with distance. The relationship among electrical force, charges, and distance, now known as Coulomb’s law, was discovered by the French physicist Charles Coulomb (1736–1806) in the eighteenth century. Coulomb’s law states that for charged particles or objects that are small compared with the distance between them, the force between the charges varies directly as the product of the charges and inversely as the square of the distance between them.

Coulomb’s law can be expressed as

$\ F = k \frac {q_{A}q_{B}}{d^2}$

where $\ d$ is the distance between the charged particles; $\ q_{A}$ represents the quantity of charge of one particle and $\ q_{B}$the quantity of charge of the other particle; and $\ k$ is the proportionality constant.

When the charges are measured in coulombs, the distance is measured in meters, and the force is measured in newtons, the constant, K, is $\ 9.0 X 10^9 N m^2/C^2$.
This equation gives the magnitude of the force that charge $\ q_{A}$ exerts on $\ q_{B}$ and also the force that $\ q_{B}$ exerts on $\ q_{A}$. These two forces are equal in magnitude but opposite in direction. You can observe this example of Newton’s third law of motion in action when you bring two strips of tape with like charges together. Each exerts forces on the other. If you bring a charged comb near either strip of tape, the strip, with its small mass, moves readily. The acceleration of the comb and you is, of course, much less because of the much greater mass.
The electrical force, like all other forces, is a vector quantity. Force vectors need both a magnitude and a direction. However, Coulomb’s law will furnish only the magnitude of the force. To determine the direction, you need to draw a diagram and interpret charge relations carefully.
Consider the direction of force on a positively charged object called A. If another positively charged object, B, is brought near, the force on A is repulsive. The force, $\ F_{B}$ on A, acts in the direction from B to A, as shown in Figure 1a. If, instead, B is negatively charged, the force on A is attractive and acts in the direction from A to B, as
shown in Figure 1b.

FIGURE 1 The rule for determining direction of force is like charges repel, unlike charges attract.

The superposition principle
• The electric force obeys the superposition principle.
• That means the force two charges exert on a third force is just the vector sum of the forces from the two charges, each treated without regard to the other charge.
• The superposition principle makes it mathematically straightforward to calculate the electric forces exerted by distributions of electric charge.
• The net electric force is the sum of the individual forces.

The Electric Field
The electric field is a vector quantity that relates the force exerted on a test charge to the size of the test charge. How does this work? An electric charge, q, produces an electric field that you can measure. This is shown in Figure 2. First, measure the field at a specific location.

FIGURE 2 Arrows can be used to represent the magnitude and direction of the electric field about an electric charge at various locations.

Call this point A. An electrical field can be observed only because it produces forces on other charges, so you must place a small positive test charge, $\ q′$, at A. Then, measure the force exerted on the test charge, $\ q′$, at this location.
According to Coulomb’s law, the force is proportional to the test charge. If the size of the test charge is doubled, the force is doubled.
Thus, the ratio of force to test charge is independent of the size of the test charge. If you divide the force, F, on the test charge, measured at point A, by the size of the test charge, $\ q′$, you obtain a vector quantity, $\ F/q$′. This quantity does not depend on the test charge, only on the charge q and the location of point A. The electric field at point A, the location of $\ q′$, is represented by the following equation.

Electric Field E

$\ E = \frac {F_{on q'}}{q'}$
$\ E = k\frac {q}{r^2}$.....[single point charge q]

The direction of the electric field is the direction of the force on the positive test charge. The magnitude of the electric field is measured in newtons per coulomb, N/C.
A picture of an electric field can be made by using arrows to represent the field vectors at various locations, as shown in Figure 2.

The length of the arrow will be used to show the strength of the field. The direction of the arrow shows the field direction. To find the field from two charges, the fields from the individual charges are added vectorially. A test charge can be used to map out the field resulting from any collection of charges.

Superposition principle of electric fields
The superposition principle states that, for a collection of charges, the total electric field at a point is simply the summation of the individual fields due to each charge

 at any point P, the total electric field due to a group of charges equals the vector sum of the electric fields of the individual charges.

The dipole: an important charge distribution
• An electric dipole consists of two point charges of equal magnitude but opposite signs, held a short distance apart.

## Waves transfer energy but not matter

#### Quiz

Q1) A wave can make a leaf bob up and down on the water, but it cannot move the leaf toward the shore. This is because waves only transfer…

A. Media
B. Crests
C. Energy
D. Matter

Q2) How do waves transfer energy?

A. By a combination of waves
B. By the vibration of particles in a medium
C. Through ocean waves
D. Through compression

#### Waves transfer energy but not matter Simulation(Virtual Experiment)

When the next simulation is not visible, please refer to the following link.

Wave motion
Consider what happens to the surface of a pond when you drop a pebble into the water. The disturbance created by the pebble generates water waves that travel away from the disturbance, as seen in Figure 1. If you examined the motion of a leaf floating near the disturbance, you would see that the leaf moves up and down and back and forth about its original position. However, the leaf does not undergo any net displacement from the motion of the waves.

Figure 1. A pebble dropped into a pond creates ripple waves similar to those shown here.

The leaf ’s motion indicates the motion of the particles in the water. The water molecules move locally, like the leaf does, but they do not travel across the pond. That is, the water wave moves from one place to another, but the water itself is not carried with it.

Ripple waves in a pond start with a disturbance at some point in the water. This disturbance causes water on the surface near that point to move, which in turn causes points farther away to move. In this way, the waves travel outward in a circular pattern away from the original disturbance.

In this example, the water in the pond is the medium through which the disturbance travels. Particles in the medium—in this case, water molecules—move in vertical circles as waves pass. Note that the medium does not actually travel with the waves. After the waves have passed, the water returns to its original position.

Waves of almost every kind require a material medium in which to travel. Sound waves, for example, cannot travel through outer space, because space is very nearly a vacuum. In order for sound waves to travel, they must have a medium such as air or water.Waves that require a material medium are called mechanical waves.

Not all wave propagation requires a medium. Electromagnetic waves, such as visible light, radio waves, microwaves, and X rays, can travel through a vacuum. You will study electromagnetic waves in later chapters.

Waves transfer energy
When a pebble is dropped into a pond, the water wave that is produced carries a certain amount of energy. As the wave spreads to other parts of the pond, the energy likewise moves across the pond. Thus, the wave transfers energy from one place in the pond to another while the water remains in essentially the same place. In other words, waves transfer energy by the vibration of matter rather than by the transfer of matter itself. For this reason, waves are often able to transport energy efficiently.

The rate at which a wave transfers energy depends on the amplitude at which the particles of the medium are vibrating. The greater the amplitude, the more energy a wave carries in a given time interval. For a mechanical wave, the energy transferred is proportional to the square of the wave’s amplitude.

When the amplitude of a mechanical wave is doubled, the energy it carries in a given time interval increases by a factor of four. Conversely, when the amplitude is halved, the energy decreases by a factor of four.

As with a mass-spring system or a simple pendulum, the amplitude of a wave gradually diminishes over time as its energy is dissipated. This effect, called damping, is usually minimal over relatively short distances. For simplicity, we have disregarded damping in our analysis of wave motions.