PHYSICS 5A Lecture Notes - Lecture 6: Projectile Motion

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2 Feb 2018

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Projectile motion: to calculate the highest point of a projectile, you use vyf=0, where t0 us the time to that point, and you can use equation vf=vi+a t where a is g to solve for vi. Then you can use xf=xi t+(1/2)ax t2 to get to the height. Using sinx to calculate the velocity in the x direction: to calculate the range (in the x direction) we use the time value calculated before, and plug it into the equation xf=xi t+(1/2)ax t2 . Calculate the speed and centripetal acceleration of the earth in its orbit around the sun. Assuming a circular orbit, r=1. 5x1011 and t=2. 15x107s v=2. 0x104 and a=6x10-3. It moves straight up to a maximum height, then falls to the ground. Which graph represents the vertical velocity of the arrow as a function of time. The answer looks as so because acceleration is constant at negative g. The diagram shows three points of a motion diagram.

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which of the vectors below best represents the direction of the impulse vector j⃗ ?

To learn about the impulse-momentum theorem and its applications in some common cases.

Using the concept of momentum, Newton's second law can be rewritten as

?F? =dp? dt, (1)

where ?F? is the net force F? net acting on the object, and dp? dt is the rate at which the object's momentum is changing.

If the object is observed during an interval of time between times t1 and t2, then integration of both sides of equation (1) gives

?t2t1?F? dt=?t2t1dp? dtdt. (2)

The right side of equation (2) is simply the change in the object's momentum p2??p1?. The left side is called the impulse of the net force and is denoted by J? . Then equation (2) can be rewritten as

J? =p2??p1?.

This equation is known as the impulse-momentum theorem. It states that the change in an object's momentum is equal to the impulse of the net force acting on the object. In the case of a constant net force F? net acting along the direction of motion, the impulse-momentum theorem can be written as

F(t2?t1)=mv2?mv1. (3)

Here F, v1, and v2 are the components of the corresponding vector quantities along the chosen coordinate axis. If the motion in question is two-dimensional, it is often useful to apply equation (3) to the x and y components of motion separately.

( The following questions will help you learn to apply the impulse-momentum theorem to the cases of constant and varying force acting along the direction of motion. First, let us consider a particle of mass m moving along the x-axis. The net force F is acting on the particle along the x-axis. F is a constant force.)

A) The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0.

B) The particle starts from rest at t=0. What is the magnitude v of the velocity of the particle at time t? Assume that t>0.

C) The particle has the momentum of magnitude p1 at a certain instant. What is p2, the magnitude of its momentum ?t seconds later?

D) The particle has the momentum of magnitude p1 at a certain instant. What is v2, the magnitude of its velocity ?t seconds later?

E) Find the magnitude of the impulse J delivered to the particle.

F) Which of the vectors below best represents the direction of the impulse vector J??
(Figure 1)

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puceworm982

Learning Goal: Understand how to determine the constants in thegeneral equation for simple harmonic motion, in terms of giveninitial conditions.
A common problem in physics is to match the particular initialconditions - generally given as an initial position x_0 andvelocity v_) at t=0 - once you have obtained the general solution.You have dealt with this problem in kinematics, where theformula

1. x(t) = x_0 + v_0t + at^2

has two arbitrary constants (technically constants of integrationthat arise when finding the position given that the acceleration isa constant). The constants in this case are the initial positionand velocity, so "fitting" the general solution to the initialconditions is very simple.

For simple harmonic motion, it is more difficult to fit the initialconditions, which we take to be

x_0, the position of the oscillator at t = 0, and
v_0, the velocity of the oscillator at t = 0.

There are two common forms for the general solution for theposition of a harmonic oscillator as a function of time t:

2. x(t) = A\cos(omega t + phi) and
3. x(t) = C\cos(omega t) + S\sin(omega t),

where A, phi, C, and S are constants, omega is the oscillationfrequency, and t is time.

Although both expressions have two arbitrary constants--parametersthat can be adjusted to fit the solution to the initialconditions--Equation 3 is much easier to use to accommodate x_0 andv_0. (Equation 2 would be appropriate if the initial conditionswere specified as the total energy and the time of the first zerocrossing, for example.)

Find C and S in terms of the initial position and velocity of theoscillator.
Give your answers in terms of x_0, v_0, and omega. Separate youranswers with a comma.

The figure shows three graphs, A, B, C, which collectively describe the motion of an object in one dimension. The horizontal axis of each graph is time (symbol t). The vertical axes of the graphs are different. For one graph the vertical axis is position (symbol x), for another, it is the velocity (symbol v), and for another, it is acceleration (symbol a). Where the axes across is the point of zero for both time and the quantity corresponding to the vertical axis. Part A: For which sequence of graphs is the vertical coordinate position, velocity, and acceleration, respectively? (e.g., If you think graph Y is position versus time, graph X is velocity versus time, and graph Z is acceleration versus time, then your answer is Y, X, Z.). Multiple choice: 1)C, A, B 2)A, C, B 3)C, B, A 4)B, C, A 5)B, A, C 6)A, B, C. Part (b) Which one of the following situations can be most accurately represented by the graph, consistent with your answer to part (a)? (1) A car driving in the negative x-direction suddenly shifts into reverse and, with tires spinning, slows to a momentary stop before beginning to move in the position x-direction. (2)A car driving in the position x-direction slams on its brakes and comes to rest. (3) A rock is dropped from the rest from the top of a tall building. Assume air drag can be neglected, in which case its speed increases by the same amount each second. Take upward to be the position x-direction. (4) A sprinter runs a 100-meter dash and she runs in the negative x-direction. Starting from rest, she has a constant acceleration for the first 50 meters, then a constant speed for the remaining 50. Multiple choices: 1)4, 2)3, 3)2, 4)1.

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PHYSICS 5A Lecture Notes - Lecture 6: Projectile Motion

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