Newton's third law of motion ( Action and Reaction ! )

When a bullet is fired from a gun equal and opposite forces are exerted on the bullet and the gun during the time the bullet is passing down the barrel

Newton's second law of motion ( momentum , force , weight , free falling )

A motortruck requires a larger force to set in motion when it is heavily loaded than when it is empty

The nature of friction: Static and dynamic friction

Friction is the name given to the force which opposes the relative sliding motion of two surfaces in contact with one another. It plays a notable part in our daily lives. For example

Newton's Laws of motion:1- Newton's first law of motion

In 1687 Newton published a book written , as was the routine in those days , in Latin and given the title , Philosophiae naturalis principia mathematica Translated , this means

Derivation of Kinetic energy formula and worked examples

Where there are no opposing forces, a moving body needs noforce to keep it moving with a steady velocity ( Newton's first law of motion ). If, however, a resultant force does act on a moving body

Tuesday, September 2, 2014

How To Solve Physics Problems Forces (including friction) problems and solutions

Forces (including friction)

In the development of mechanics, the first thing to learn is the interrelation of position, velocity, and acceleration (for constant acceleration). These interrelations are described with the four kinematic equations of motion and discussed in earlier chapters.

In this chapter we relate constant force to constant acceleration and then to all of kinematics. Operationally, force can be defined as what makes masses accelerate. Actually it is the unbalanced force on a mass that makes it accelerate. Simply: Unbalanced forces make masses accelerate.

Force, acceleration, and mass are related by Newton's second law as F = ma.

5 1 A 40 N force is applied to a 20kg block resting on a horizontal frictionless table. Find the acceleration?

Solution: The acceleration is

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Fig. 5 1

Once the acceleration is known, the kinematic equations of motion can be applied to find position, velocity, and acceleration as a function of time.

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5 2 Now place the 20kg mass (with no external force applied) on a frictionless 50° incline plane. What is the acceleration of the mass?

Solution: The force acting on the mass is due to gravity so set up a vector diagram (some authors call this a force diagram or free body diagram) starting with this 196N force acting down. The mass is constrained to move down the plane, so the force we want is the component of this gravitational force acting down the plane. In Fig. 5 2 the gravitational force is shown with components down the plane, Fp, and normal, Fn, to the plane. Notice the geometry of the situation where the 50° angle of the plane is the same as the angle between the gravitational force and the normal (force).

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The 150N force acting down the plane causes the 20kg mass to accelerate at

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5 3 Place a 10kg mass on a frictionless 35° incline plane, and attach a second 20kg mass via a cord to hang vertically as shown in Fig. 5 3. Calculate the acceleration of the system.

Solution:

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Fig. 5 3

This problem introduces the concept of the tension in the connecting cord. The most convenient way to visualize this tension is that if the cord were cut and a force meter inserted it would read a certain tension (force). This tension acts, as shown, on each mass. Notice the 35° angle of the plane and the 35° angle between the normal and the direction of mg.

We'll do the problem first with the tension and later without the tension.

To write equations relating the unbalanced force to the acceleration, we need to make an assumption as to which way the masses move. Assume that the system (of masses) moves to the right or clockwise.

Start by writing equations relating the unbalanced force on each mass. On m1 the unbalanced force is T   56N, and this force makes the 10kg accelerate.

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Likewise on m2, the unbalanced force to make the mass accelerate is

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The assumption that the system accelerates clockwise dictates the directions of the forces in these two equations. Look again at these equations and notice that if we had assumed the acceleration were counterclockwise then the signs would be different. In frictionless problems it is all right to have a negative acceleration. The negative sign means that you guessed wrong in assuming the acceleration direction. This is not true with friction problems.

Add the two equations and take the accelerations the same. The accelerations are the same because the masses are linked together with the cord.

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Now the tension in the cord is easily calculated from the equation for m1.

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Another way to view this problem is to say that the unbalanced force on m1 and m2 is (196 56)N = 140N. This force makes the total mass of 30kg accelerate, so

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This is a very quick way to find the acceleration of the system, but it is inconvenient to find the tension in the cord.

5 4 The next complication in force problems with incline planes is a double incline plane as shown in Fig. 5 4.

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Fig. 5 4

Solution: The vector diagram for these masses is shown in Fig. 5 5. Assume the system moves clockwise.

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The equation for m1 is T - 76N = 12kg.a
The equation for m2 is 83N - T = 20kg.a
Adding the two equations 7.0N = 32kg.a or a = 0.22m/s2
The tension in the cord is T = 76N + 12kg.0.22m/s2 = 79N

5 5 Another popular problem is the Atwood machine. A simple example is shown in Fig. 5–6. Find the acceleration of the system. Solution: Assume the system moves counterclockwise.

The equation for m1 is 98N–T = 10kg.a
The equation for m2 is T - 147N = 15 kg.a
Add the equations -49N = 25kg.a or a = -2.0m/s2

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The system accelerates clockwise! The tension in the cord is from the equation for m1.

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Friction
A good first problem in friction is the arrangement shown in Fig. 5–7 where the problem is first done without
friction and then with friction.
For the purposes of doing problems there are two important properties of frictional forces:
· they oppose the motion, and
· they are less than or equal to a constant (the coefficient of friction) times the normal force, the force at
the frictioning surface. (f < μN)
5–6 Two masses are arranged with one, of 50kg, on a frictionless table and the other, of 30kg, attached by a
cord and hanging, over a frictionless pulley, off the edge of the table. Find the acceleration of the system.
Solution: Assume the system moves clockwise.

The equation for m1 is

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The equation for m2 is

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Adding the two equations

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5–7 For the situation of problem 5–6 add a coefficient of friction ì = 0.20 between the block and table. Solution: This makes the vector diagram look like Fig. 5–8. Notice that the only addition is the frictional retarding force ìN = 98N.

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The equation for m1 is T - 98N = 50kg.a
The equation for m2 is 294N - T = 30kg.a
Add the equations to obtain 196N = 80kg.a or a = 2.4 m/s2

5–8 For the situation of problem 5–7 increase the coefficient of friction to 0.80. What is the acceleration of the system?

Solution: This makes the vector diagram look like Fig. 5–9.

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The equation for m2 is T - 392N = 50 kg.a
The quation for m2 is 294 - T = 30kg.a
Adding the tow equations -98N = 80kg.a or a = -1.2m/s2

Clearly the system does not accelerate counterclockwise! This is an illustration of the second property of frictional forces: the frictional force is less than or equal to ìN. In this case the frictional force just balances the tension in the cord. The frictional force, as calculated, can be up to 392N. In order for m2 not to move, the tension in the cord must be 294N. (If m2 is not moving, the forces must be in balance.) Therefore the forces on m1 must be 294N due to the tension in the cord and 294N due to friction.

Another important point with frictional forces is the normal force as illustrated by this next problem.

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5–9 A 50kg sled is pulled along a level surface at constant velocity by a constant force of 200N at an angle of 30°. What is the coefficient of friction between the sled and the surface?

Solution: The first step in this problem is to find the components of the applied force and the mg of the sled. These are shown on the vector diagram. Notice that the normal force, the actual force between the sled and the friction surface is mg minus the component of the applied force lifting up on the sled. Sometimes it is easy to miss this vertical component of the applied force. This is the major mistake in this type of problem. To help visualize this vertical component of force try placing the origin for the vectors in the middle of the sled rather than at one end where a rope would be attached.

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Fig. 5–10

The normal force is mg(=490N) minus the vertical component of the applied force, 100N, or 390N. Since the sled is moving at constant velocity the forces must be in equilibrium. The horizontal component of the applied force must equal the frictional retarding force, the coefficient of friction times the normal force.

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5–10 A 65°  incline plane has a mass of 20kg on the plane where the coefficient of friction is 0.40 and a mass of 50kg hanging free. Calculate the acceleration of the system.

Solution: Set up the vector diagram assuming motion counterclockwise. In the diagram the frictional force is calculated as a maximum of 33N.

The equation for m2 is 490N - T = 50kg.a
The equation for m1 is T - 178N - 33N = 20 kg.a
Adding the equations 279N = 70kg.a or a = 4.0m/s2

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In doing friction problems like this, you have to be careful with the frictional force. If the forces up and down the plane are nearly balanced, it is important to remember that the frictional force is not 33N but that it can be up to 33N. In a problem similar to this, it could easily happen that the blocks would not move.

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5–11 Consider a double incline plane with angles, masses, and coefficients of friction as shown in Fig. 5–12. Calculate the acceleration of the system.

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Fig. 5–12

Solution: Assume the system moves counterclockwise and set up the vector diagram.

The equation for m1 is (30 - 15 - TN = 6.0kg.a
The equation for m2 is (T - 62 - 10)N = 9.0kg.a
Adding the equations -57N = 15kg.a

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This is a negative acceleration. We cannot simply reverse the sign of a to find the correct answer since if the system were moving the other way the frictional forces would have to be reversed. Further analysis requires us to consider another vector diagram assuming the system moves in a clockwise direction.

The numbers have all been calculated so the vector diagram can be written down easily and is shown in Fig. 5–14

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The new equations are: (T - 30 - 15)N = 6.0kg.a
(62 - T - 10)N = 9.0kg.a

Adding the equations yields 7.0N = 15kg.a or a = 0.54 m/s2 The tension in the cord is from the first equation: T = 45N + 6.0kg(0.54m/s2) = 48N If the acceleration of the system were negative for both the clockwise and counterclockwise calculations then we would conclude that the system would not move. The final complication in these incline plane problems is the introduction of three masses and two connecting cords.

5–12 Consider a flat surface with a coefficient of friction of 0.20 with a 2.0kg mass and a 3.0kg mass connected together with a 6.0kg mass along a slant as shown in Fig. 5–15. The coefficient of friction on the slant is zero. Calculate the acceleration of the system.

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Fig. 5–15

Solution: In this problem note that the tensions are different leading to three unknowns, T1, T2, and a. The solution will require three equations from three vector diagrams as shown in Fig. 5–16. Assume the acceleration of the system is clockwise.

The frictional retarding force on the 2.0kg block is f = ìN = 0.20(19.6N) = 3.92N.

The frictional retarding force on the 3.0kg block is f = ìN = 0.20(29.4N) = 5.88N.

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The equations for the masses are: T2 = 3.92N = 2.0kg.a
T1 - T2 - 5.88N = 3.0kg.a
24.8N - T1 = 6.0kg.a

The last two equations added together are: 18.9N - T2 =9.0kg.a
and adding this to the first equation yields 15.0N = 11 kg.a or a = 1.36 m/s2
The tensions come from the first and third equations for the masses.

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Circular Motion A particle moving at constant speed in a circle is moving in Uniform circular motion. There is an acceleration and a force associated with such a particle. Figure 5–17 shows the velocity vectors associated with a particle in uniform circular motion.

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Fig. 5–17

While the length of the velocity vector remains the same, the direction changes continually. The diagram on the left of Fig. 5–17 shows vi and vf, with the tail of the vectors at the same point, and Äv. Looking at the limiting case where Ä è 0 the ratioimage

is valid because of similar triangles. With this statement write

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which in the limiting situation when Äè 0 or Ä t 0 (same criterion) is

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The “rad” subscript is a reminder that the acceleration points radially inward. The direction of arad is also clear from the limiting situation. As Ä t 0 the angle between v and Ä v 90°, so the acceleration is directed radially inward.

5–13 A certain car driven in a circle can exert a maximum side force of 0.95g. What is the maximum speed for this car driven in a circle of 160m radius?

Solution: An 0.95g means a side force or acceleration directed at right angles to the direction of travel of 0.95(9.8m/s2) = 9.3m/s2. Using arad = v2/r

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At a speed greater than 38.6m/s the car will “slide out of the circle.”

The acceleration directed toward the center of the circle for a mass in unifom circular motion is called the centripetal acceleration. The force associated with moving a mass in a circle is marad and is called the centripetal force.

5–14 What is the (centripetal) force produced by the tires acting on the pavement for the car of problem 5–13 if the car is 1200kg of mass?

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Solution:

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5–15 An 0.60kg rubber stopper is whirled in a horizontal circle of 0.80m radius at a rate of 3.0 revolutions per second. What is the tension in the string?

Solution: Three revolutions per second means three 2 ðr's (circumferences) per second so

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The tension in the string, which is providing the centripetal force, is

image

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5–16 A 1 oz gold coin is placed on a turntable turning at 33 1/3 RPM (revolutions per minute). What is the coefficient of friction between the coin and the turntable if the maximum radius, before the coin slips, is 0.14m?

Solution: The frictional force between the coin and turntable provides the center directed force to keep the coin on the turntable (see Fig. 5–18). This center directed force must equal my2 / r.

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Fig. 5–18

This is a force balance problem. In equation form: frictional force = centripetal force, or

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The velocity is 2ðr times the number of 2 ðr's per minute, 100 / 3, or

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Therefore,image

5–17 A conical pendulum is a mass on the end of a cord, where the mass moves at constant speed in a circlewith the cord tracing out a cone. A conical pendulum of length 1.2m moves in a circle of radius 0.20m. What isthe period of the pendulum?
Solution: Note that the mass is not given. Start with a vector diagram of the forces on the mass as shown in Fig.
5–19.
The mg must equal the vertical component of the tension in the string T cosθ = mg.
The horizontal component is due to the centripetal force so T sin θ = mv2/r.
Dividing the second equation by the first eliminates T and m or tan θ = v2/rg.
The angle is from sin θ = r/L making tan θ = 0.17 and

image

image

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The period is from 2πr/T = v or
5–18 What is the speed, at which no side force due to friction is required, for a car traveling at a radius of 240m on a 20° banked road?
Solution: Figure 5–20 shows a profile of the car on the banking and the vector diagram.

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Start with mg acting down. The normal force (between the car and road) is normal to the surface with the vertical component equal to mg and the horizontal component equal to mv2/r.

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And again,

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so,

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Notice that in this problem the mass of the car does not enter into the calculation.

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Go back over problems 5–13 through 5–18 and notice how each problem could be written asking for a differentvariable. For example in problem 5–13 give the side force and the velocity and ask for the radius. In problem 5–16 give the speed and the coefficient of friction and ask for the radius. Changing the problems like this is anxcellent way to generate practice (for the test) problems.

How To Solve Physics Problems Projectile Motion problems and solutions

Projectile Motion

In order to understand projectile motion you have to look at the motion in two directions with one direction oriented in the direction of constant acceleration and the other direction at a right angle to it so as to form an x y coordinate system. In most problems, where an initial velocity and angle with the horizontal is given, the velocity is written in component form.

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Fig.4 1

Note that the motion is effectively separated into horizontal and vertical components (one in the direction of the constant acceleration and one at a right angle to the acceleration).

The motion is a parabola in this coordinate system. The symmetry of parabolas (in this case parabolic motion) is helpful in understanding the motion. (See the Introduction, Mathematical Background, for a discussion of the properties of parabolas.) In order to better understand how the motion is separated and how the parabolic property helps us to understand projectile problems we will do a simple problem without numbers just to get a feel for the features of the motion. If you are familiar with the properties of parabolas you may want to skip this discussion and go directly to the worked problems.

In Fig. 4 1 the velocity of the projectile is written in component form. The acceleration in the horizontal direction is zero and in the vertical direction is due to gravity. With the velocity components, and the acceleration in the direction of an axis, we can write the six equations describing acceleration, velocity, and position in the x and y directions.

These equations are based on the kinematic equations of motion for constant acceleration:

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The six equations are:

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The equations for x and y can be looked upon as parametric equations in time. Parametric equations such as x = f(t) and y = f (t) are equations that can be combined to produce y = f (x) or x = f (y). To find the position of the particle in x–y, or y as a function of x, solve one equation for t and substitute into the other. In this case solve x=… for t, as this is the simplest choice.

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Substitute into y = … to obtai

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This is of the form y =  ax2 + bx = x( ax+b), which is a parabola that opens down and intercepts the y axis (makes y = 0) at x = b/a.

The range, or value of xwhen y = 0, can be determined from the factored form of this equation

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which tells us that y = 0 at x = 0 and

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or using the trigonometric identity 2sinθcosθ=sin2θimage

The maximum range occurs for an angle of 45°, corresponding to sin 2è= 1.

The main point of this discussion is that the motion is a parabola, and the properties of parabolas can be used in solving problems in projectile motion. That the maximum range occurs at 45° is not surprising, and is generally not of interest in problems.

The expression for maximum range is only of passing interest, since the range is one of the easier things to calculate in any problem. Do not consume precious memory space memorizing formulas for the range or time of flight. If you work the problems by first writing down the six equations describing the motion, then the time of flight, range, and many other things are easily calculated.

In order to set up a problem in projectile motion, first orient one axis of a right angle (x–y) coordinate system in the direction of constant acceleration. Remember to place the origin of the coordinate system and the positive direction for x and y for (your) convenience in solving the problems. Then write equations for acceleration, velocity, and position for each direction. Along with the symmetry of the motion, these equations can be used to find the characteristics (position, velocity, and acceleration) of the projectile at any point in space or time. The procedure is illustrated with a simple problem.

4 1 A soccer player kicks a ball at an initial velocity of 18m/s at an angle of 36° to the horizontal. Find the time of flight, range, maximum height, and velocity components at t = 0, mid range, and at impact.

Solution: The acceleration is down so set up the coordinate system with x horizontal (along the ground ) and y vertical, and place v0 on the graph. Write the velocity in component form, and calculate vx0 and vy0.

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Now, and this and this is the key to getting the problem right, write down the six equations governing the motion starting with the accelerations (the accelerations are the easiest to write).

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Now look at the motion, which is parabolic in x–y, keeping in mind the properties of parabolas.

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Fig.4 3

The ball is on the ground at t = 0 and t = ô, the time of flight. To find these times, set y = 0 and obtain (without units)

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which gives t = 0 and t = 10.6/4.9 = 2.16s, the time of flight.

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The range is the value of x at 2.16s:

The maximum height, and vy = 0, occur at 1.08s (one half the time of flight). Therefore the maximum height is y at t = 1.08s

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The velocity components are:image

The procedure for doing the problem is to write the initial velocity in component form, write the six equations for acceleration, velocity, and position, and then perform the mathematical operations answering questions about the problem.

4 2 For the situation of problem 4 1 suppose the field is covered with fog down to 3.0m above the ground. What are the times for the ball's entry into and exit out of the fog?

Solution: The times are found from the equation for y as a function of time with y set equal to 3.0m.

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As an exercise find the x positions for these times.

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4 3 An airplane traveling at 100m/s drops a bomb from a height of 1500m. Find the time of flight, distance traveled, and the velocity components as the bomb strikes the ground.

Solution: Start by placing the origin of the coordinate system at the point where the bomb is released. And take the direction the bomb falls due to gravity as positive x and the horizontal position as y. This is different from the conventional orientation, but it is convenient in this problem because the six equations all come out positive!

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Now write down the six equations governing the motion.

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With these six equations we can answer all the questions in the problem. We need only to “translate” the word questions into algebra questions.

a) The time for the bomb to reach the earth means: Find the time when x = 1500m?

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b) How far does the bomb travel horizontally means: Find the value of y when t = 17 s?

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c) Find the velocity components at impact means: Find vx and vy at t = 17s?

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d) Where is the airplane when the bomb strikes the earth means: What is y and t = 17 s?

Remember that the plane and the bomb have the same velocity in the horizontal direction.

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4 4 A baseball is hit at a 45° angle and a height of 0.90m. The ball travels a total distance of 120m. What is the initial velocity of the ball? What is the height of the ball above a 3.0m fence 100m from where the ball is hit?

Solution: This problem is unique in that it requests the initial velocity, a number usually given in the problem. Also note that the angle is 45°, the angle for maximum range. The other interesting feature of the problem is the question concerning the height at some specific point down range. Set up the problem in the conventional way assigning v0 to the initial velocity.

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The next question is how to handle the ball being hit at 0.90m above the ground. Since v0 is specified at the point where the ball is hit, and this is where we start counting time, then the origin for the coordinate system should be put here. We just need to keep in mind that this origin is 0.90m above the ground level.

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Now write the six equations.

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The first question reduces to finding vo for the 120m hit. Algebraically this means that when y =  0.9m, x = 120m, so write the two equations (for x and y) with these conditions

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Note that so write

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This is the time of flight, so put this time into the equation for x.image

The next part of the problem asks for the height above the fence at 100m. Again we need to find the time for x to be 100m down range and substitute this value of t into the y equation

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This is the time for the ball to go 100m. The height of the ball at this time isimage

Remember that this is 15.6m above the zero of the coordinate system which is 0.90m above the ground. So the ball is 16.5m above the ground at this point and for a 3.0m high fence is 13.5m above the fence.

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4–5 A physicist turned motorcycle stunt rider will jump a 20m wide row of cars. The launch ramp is 30° and 9.0m high. The land ramp is also 30° and is 6.0m high. Find the minimum speed for the launch.

Solution:

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Fig. 4–7

First write down the six equations of motion.

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The minimum v0 is dictated by the condition that the rider be at x = 20m and y = 3.0m. So put these conditions into the equations for x and y.

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At this point we can solve for v0 by substituting

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from the first equation into the second equation.

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This is the minimum velocity to make the jump.

As a follow on to this problem calculate the velocity to just hit the end of the ramp. This velocity gives an upper limit to the jump velocity.

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Go back through the problems and notice the common approach:

1) orient a coordinate system,

2) write the six equations of motion, and

3) translate the word questions into algebra.

In these problems we have not always used each of the six equations to answer the question. Few problems require the detailed description of the motion available in the six equations. If, however, each problem is described by these six equations you will always have all the information to answer any possible question about the motion.