PHYS 205 Final: PHYS 206 TAMU Final Exam 2009 Solutions

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turquoisegnu128

31 Jan 2019

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<1: find the angular momentum of the plane about the origin as a function of time. d. A certain spring is found not to obey hooke"s law; it exerts a restoring force fjx) = ,&t f it is stretched or compressed, where a and /1 are known constants. The mass of the spring is negligible: calculate the potential energy function u(x) for the spring. n. I z: an object with mass m on a frictionless, horizontal surface attached to this spring is pulled a a distance a to the left (in the x direction) and then released. If the surface is frictionless, find the velocity of the object when it is a12 to the right of the equilibrium position. x= a x=o f. I: how the answer in b) will change is there is a coefficient of friction between the block and the surface, t. (assume that the block will reach this point)

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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

1. The Yukawa Potential energy function for the interaction ofneutrons and protons in a nucleus is . (a) What is the forceassociated with this potential energy? (b) If and Find the value ofthe force at r=a and at r=3a

2. For what value of the constant c is the forceconservative?

3. In a two-stage rocket, the exhaust velocity for the first stageburn is 3000 m/s and for the second stage burn is 3000 m/s. Assumethat during the entire sequence of burns the rocket is still nearenough to the surface of the earth to approximate the externalgravitational acceleration as 9.8m/s^2 , and ignore air resistance.The mass of the first stage engine and tanks is 2000kg. The mass ofthe fuel held in the first stage is 4000kg. The first stage fuelburns in 10 seconds. The first stage is then ejected. The mass ofthe second stage engine, tanks, and payload is 2000kg. The mass ofthe fuel held in the second stage is 2000kg. The second stage fuelburns in 5 seconds. (The second stage tanks remain attached to thepayload when empty.) (a) At what speed is the second stage/payloadcombination moving at the end of the second stage burn? (b) Howmuch faster is this than the speed if all the 6000kg fuel wereburned in the first stage in 15 seconds with the same exhaustvelocity (3000m/s) and the entire 4000kg assembly of the emptyrocket remained in one piece?

4. A particle with mass 10kg starts from rest at r = 10 and ?=0radians, following the trajectory given by r=10-2t , ?=0.2t. Findthe radial and angular components and of the force, as functions oftime, that will cause this motion

5. A sheet of material, mass 2kg, uniform density, is in the shapeof a right triangle bounded by the lines y=0, x=0, and y=4-x. Itattched to the origin, and free to rotate about the y axis. A forceof 1000 Newtons is applied perpendicular to the plane, located atthe x=4 point on the triangle as it starts to rotate. In aright-handed coordinate system, how fast will the object bespinning (in radians per second) as it crosses the y-z plane (thatis, when it has rotated through 270 degrees)?

Thank you in advanced.

Figure 1 illustrates a Keplerian orbit, with Cartesiancoordinates (x,y) and polar coordinates (r,Ï).

Parametric equations for a general Keplerian orbit are r(Ï) = a(1 â e cosÏ)

t(Ï) = *(T)/2Ï+ ( Ïâ e sinÏ)

tan(Ï/2) = *(1+e)/(1âe)+1/2 tan(Ï/2)

Here Ï is an independent variable. The solution is periodic in Ïwith period âÏ = 2Ï. Also, a, e, T are constants that determine theparameters of the orbit.

Figure 2 shows the relation between Ï and the spatialcoordinates. In the figure, the coordinates (Î¾,Î·) are defined by Î¾= x + ae and Î· = y a /b,

withb=a*1âe2 ]1/2.

(A) What kind of curve is the orbit?

(D) Determine Î¾(Ï) and Î·(Ï).

(E) Express r as a function Ï .

(F) The angular momentum is L = m r2dÏ/dt. Determine L in termsof {a,e,T} Hence verify that L is a constant of the motion.

(G) The energy isE = [1/2] m ( dr/dt )2+ [1/2] m r2(dÏ/dt)2â GMm/r.Determine E in terms of {a,e,T}.

(H) E must be a constant of the motion. Hence determine T and Efrom the equation that you obtained in (F).

(I) Write T in terms of the spatial orbit parameters and theforce constant GM.

(J) Now express L and E in terms of a, e, GM.

(K) In Figure 1, determine the coordinates of the points P, A,and R

(L) In Figure 1, determine the time t at the point R.

PHYS 205 Final: PHYS 206 TAMU Final Exam 2009 Solutions

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