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.

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

Answer) E.

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

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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

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).

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.

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
Answer) B.

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

Answer) D.
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 momentumgives

$\ 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.

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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.

Related Videos(Waves transfer energy but not matter)

Q1) A point on a standing wave that is always stationary is a

A. maximum
B. minimum
C. node
D. antinode

Answer) C.

Q2) Standing waves are produced by the superposition of two waves with
A. the same amplitude, frequency, and direction of propagation.
B. the same amplitude and frequency, and opposite propagation directions.
C. the same amplitude and direction of propagation, but different frequencies.
D. the same amplitude, different frequencies, and opposite directions of propagation.

Answer) B.

Standing Waves on a String Simulation

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Standing waves
If you tie a rope to a wall and shake the free end up and down, you will produce a wave in the rope. The wall is too rigid to shake, so the wave is reflected back along the rope to you. By shaking the rope just right, you can cause the incident (original) and reflected waves to form a standing wave. A standing wave is a wave that appears to stay in one place—it does not seem to move through the medium.

Certain parts of a standing wave remain stationary. Nodes are the stationary points on a standing wave. Interestingly enough, you could hold your fingers on either side of the rope at a node, and the rope would not touch them. Other parts of the rope would make contact with your fingers. The positions on a standing wave with the largest amplitudes are known as antinodes. Antinodes occur halfway between nodes.

FIGURE 1. The incident and reflected waves interfere to produce a standing wave. The nodes are places that remain stationary.

Standing waves are the result of interference. When two waves of equal amplitude and wavelength pass through each other in opposite directions, the waves are always out of phase at the nodes. As Figure 1 shows, the nodes are stable regions of destructive interference.

Standing waves have nodes and antinodes

Figure 2(a) shows four possible standing waves for a given string length. The points at which complete destructive interference happens are called nodes. There is no motion in the string at the nodes. But midway between two adjacent nodes, the string vibrates with the largest amplitude. These points are called antinodes.
Figure 2(b) shows the oscillation of the second case shown in Figure 2(a) during half a cycle. All points on the string oscillate vertically with the same frequency, except for the nodes, which are stationary. In this case, there are three nodes (N) and two antinodes (A), as illustrated in the figure.

Figure 2. (a) This photograph shows four possible standing waves that can exist on a given string. (b) The diagram shows the progression of the second standing wave for one-half of a cycle.

In general, the motion of an oscillating string fixed at both ends is described by the superposition of several normal modes. Exactly which normal modes are present depends on how the oscillation is started. For example, when a guitar string is plucked near its middle, the modes shown in Figure 3(b) and (d), as well as other modes not shown, are excited.

Figure 3. (a) A string of length L fixed at both ends. The normal modes of vibration form a harmonic series: (b) the fundamental, or first harmonic; (c) the second harmonic; (d) the third harmonic.

In general, we can describe the normal modes of oscillation for the string by imposing the requirements that the ends be nodes and that the nodes and antinodes be separated by one fourth of a wavelength. The first normal mode, shown in Figure 3(b), has nodes at its ends and one antinode in the middle. This is the longestwavelength mode, and this is consistent with our requirements. This first normal mode occurs when the wavelength $\lambda _{1}$ is twice the length of the string, that is, $\lambda _{1}=2L$.

The next normal mode, of wavelength $\lambda _{2}$ (see Fig. 3(c)), occurs when the wavelength equals the length of the string, that is, $\lambda _{2}=L$.

The third normal mode(see Fig. 3(d)) corresponds to the case in which $\lambda _{3}=2L/3$.

In general, the wavelengths of the various normal modes for a string of length $\ L $ fixed at both ends are

where the index $\ n$ refers to the nth normal mode of oscillation. These are the possible modes of oscillation for the string. The actual modes that are excited by a given pluck of the string are discussed below.

The natural frequencies associated with these modes are obtained from the relationship $\ f=v/\lambda$, where the wave speed $\ v$ is the same for all frequencies. Using Equation (1), we find that the natural frequencies $\ f_{n}$ of the normal modes are

In motion of a body thrown in horizontal direction, which of these is true about the components of velocity of the system? (Assume system is a ground projectile.)

a. Only vertical component is constant

b. Only horizontal component is constant

c. Both horizontal and vertical components are constant

d. None of the components is constant

Answer) b.

Motion of a body thrown in horizontal direction : a combination of horizontal motion with constant horizontal velocity and vertical motion with a constant downward acceleration due to gravity.

Motion of a body thrown in horizontal direction Simulation

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Projectile motion is free fall with an initial horizontal velocity

To understand the motion a projectile undergoes, first examine Fig 01. The red ball was dropped at the same instant the yellow ball was launched horizontally.

Fig. 01 This is a strobe photograph of two table-tennis balls released at the same time. Even though the yellow ball is given an initial horizontal velocity and the red ball is simply dropped, both balls fall at the same rate.

The curved path shown in Fig 01. is the combination of constant horizontal motion and vertical motion that undergoes acceleration due to gravity.

If air resistance is disregarded, both balls hit the ground at the same time. By examining each ball’s position in relation to the horizontal lines and to one another, we see that the two balls fall at the same rate. This may seem impossible because one is given an initial velocity and the other begins from rest. But if the motion is analyzed one component at a time, it makes sense.

First, consider the red ball that falls straight down. It has no motion in the horizontal direction. In the vertical direction, it starts from rest ($\ v_{0y}$ = 0 m/s) and proceeds in free fall. Thus, the kinematic equations from the chapter “Motion in One Dimension” can be applied to analyze the vertical motion of the falling ball, as shown on the next page. Note that on Earth’s surface the acceleration ($\ a_y$) will equal $\ g$ ($\ 9.8 m/s^2$) because the only vertical component of acceleration is free-fall acceleration.

VERTICAL MOTION OF A PROJECTILE THAT FALLS FROM REST

$\ a_y = g$ $\ v_y = gt$ $\ y = \frac{1}{2}gt^2$

Now consider the components of motion of the yellow ball that is launched in Fig 01. This ball undergoes the same horizontal displacement during each time interval. This means that the ball’s horizontal velocity remains constant (if air resistance is assumed to be negligible). Thus, when the kinematic equations are used to analyze the horizontal motion of a projectile, the initial horizontal velocity is equal to the horizontal velocity throughout the projectile’s flight. A projectile’s horizontal motion is described by the following equation.

HORIZONTAL MOTION OF A PROJECTILE

$\ v_x = v_{0} =constant$ $\ x = v_{0}t $

Next consider the initial motion of the launched yellow ball in Fig 01. Despite having an initial horizontal velocity, the launched ball has no initial velocity in the vertical direction. Just like the red ball that falls straight down, the launched yellow ball is in free fall. The vertical motion of the launched yellow ball is described by the same free-fall equations. In any time interval, the launched ball undergoes the same vertical displacement as the ball that falls straight down. For this reason, both balls reach the ground at the same time.

To find the velocity of a projectile at any point during its flight, find the vector that has the known components. Specifically, use the Pythagorean theorem to find the magnitude of the velocity, and use the tangent function to find the direction of the velocity.