PHYS 150

Dr. Johnny B. Holmes

Introduction
Work, Energy and Power
Linear Momentum and Collisions
Return to PHYS 150 Outline

In this third quarter of the course we consider: (a) another fundamental law of nature called conservation of energy that will help us in our consideration of motion and which involves the concepts of work and energy; and (b) the concept of momentum which is especially useful in considering collisions.

In the first part we introduce the concepts of WORK and ENERGY. We look at two particular kinds of energy: kinetic (or motional) energy and potential energy (energy due to position). We also look at one of the most powerful tools in physics: THE LAW OF CONSERVATION OF ENERGY. Finallly, we consider the related concept of POWER.

In the second part we introduce the concept of MOMENTUM and apply it to collisions. In so doing we develop the law of conservation of momentum.

Return to top

Outline
Supplementary Homework Problems
Answers to Supplementary Problems
Return to top

1. Work
1. definition: exert a force through a distance
2. units: Joules
3. dot product: W = F· s = F s cos(q _Fs)
4. most general formula (when force changes over the distance): W = _siò ^sf F cos(q _Fs) ds
2. Energy
1. definition: the capacity to do work (under ideal conditions)
2. units: Joules
3. kinetic energy: KE = (1/2)mv²
4. potential energy (energy due to position): D PE = PE(s_f) - PE(s_i) = - _siò ^sf F cos(q _Fs) ds
1. gravitational potential energy: PE(h) = mgh (near earth)
   or, PE(r) = -Gm₁m₂/r (everywhere)
2. spring potential energy: PE(x) = (1/2)kx²
3. Conservation of Energy
1. without frictional losses: KE_initial + SPI_initial = KE_final + SPE_final
2. with losses due to friction: KE_initial + SPI_initial = KE_final + SPE_final + E_lost
3. with work added or subtracted: KE_initial + SPI_initial + W_on = KE_final + SPE_final + W_by + E_lost
4. Note: there can be more than one type of potential energy (for each particle), and there can be more than one particle (and hence more than one kinetic energy).
4. Power
1. definition: P_average = DW/Dt; P_{instantaneous} = dW/dt
2. mks units: Watts; 1 hp = 746 Watts
3. cost of utilities is proportional to Energy used (Energy = Power * time)

Return to top

30. What is the escape velocity: a) for the earth (Mass of earth is 6.0 x 1024 kg, radius is 6371 km.)? b) for the moon (Mass of moon is 7.3 x 1022 kg, radius is 1741 km.)?

31. a) Draw a graph of the potential energy due to gravity of the earth (general form), PE(r) versus r for an object of mass = 70 kg. b-1) Choose an E_total such that PE(R) < E_total < 0 and show using a solid line this constant E_total on your graph. b-2) Indicate the significance of the point of intersection of your PE(r) curve and your E_total line, and determine r and v at this intersection point. b-3) For r=R_earth , determine the PE(r), the KE(v) and v. c) Choose E_total = 0 and repeat all three steps of part b. [Note: your v for part c-3 should be the same as the escape velocity from problem 30-a.

32. a) Draw a graph of PE(x) versus x for a mass, m, of 3 kg attached to a spring where the spring constant, k is 24 Nt/m and neglect any effects of gravity. b-1) Calculate an E_total given that the velocity (v_x) at x=0 (the equilibrium point on the spring) is 5 m/s. b-2) Draw this E_total on the graph and indicate the significance of where the PE(x) curve intersects the E_total line. b-3) Determine the values for x where the PE(x) curve intersects the E_total line. c) Compare this attack of the problem and your maximum value for x with that of problem 25 from the previous part.

33. A sled of mass 5 kg is at the top of a hill 20 meters in vertical height above the base of the hill. The hill has a constant grade of 30° with the horizontal. a) If there is no friction between the sled and the ground, how fast will the sled be going when it reaches the base if it has started from rest? b) If there is a coefficient of friction between the sled and the ground of m = 0.1, how fast will the sled be going when it reaches the base if it has started with a speed of 5 m/sec? c) If the sled has a mass of 10 kg, would the answers to (a) and (b) change?

34. The average power output of the sun per unit area above the atmosphere is 1.35 kW/m². However, the average power per area reaching the ground (including day-night, clear-cloudy, summer-winter) is about 250 Watts/m². If 20 square meters of solar collectors are mounted on a house and the collectors are 25% efficient, and if the cost of energy is $.06/(Kw hr), how much money will the collectors save in one year? (There are several other factors that need to be considered to make this analysis more applicable such as storage of the energy for use when its needed and capital costs of the collector.)

35. Extra Credit: Redo problem 24, part b, but replace the constant force of 4000 Nt with a constant power (you choose the power). Be sure to include air resistance and show your logic and your results.

Return to top

30. a) 11,200 m/s = 25,000 mph; b) 2,370 m/s = 5,300 mph.

31 & 32 You are on your own on these two.

33. a) 19.8 m/s = 45 mph; b) 18.7 m/s = 42 mph; c) No.

34. $657.45

35. depends on your choice of power.

Return to top

Outline
Supplementary Homework Problems
Answers to Supplementary Homework Problems
Return to top

1. Force and momentum
1. p = mv, where p is momentum, a vector (has magnitude and direction)
2. Units: kg*m/sec (no special name for this combination)
3. Newton's Second Law: SF=dp/dt or, in component form: SF_x=dp_x/dt, SF_y=dp_y/dt
2. Impulse = Dp = ò F dt
3. Conservation of Momentum: Sp_i = Sp_f , if SF=0.
1. for two objects: m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f , a vector equation
2. work in components, so this is really two equations (in 2-D).
4. Collisions
1. elastic: E_lost = 0; bounces without denting or bruising
2. inelastic: E_lost > 0;
1. dents
2. becomes embedded: v_1f = v_2f
3. goes through
5. Rockets: conservation of momentum: v_final = v_initial + v_exhaust ln[M_initial/M_final]
   Note: M_final = M_initial - D M, where D M is the fuel burned
6. Center of Mass

Return to top

36. A constant net force of 5 Nt in the x-direction is applied to an object of mass 10 kg for 6 sec. What is (1) the total change of momentum, (2) the final velocity, and (3) the energy gain or loss of the object... a) if the object started at rest? b) if the object was going at a speed of 4 m/sec in the positive x-direction? c) if the object was going at a speed of 4 m/sec in the negative x-direction? d) if the object was going at a speed of 4 m/sec in the positive y-direction?

37. An unlucky bystander finds himself in the center of a shootout between the good guys and the bad guys. A 5.0 gm bullet moving at 100 m/sec strikes him and lodges in his shoulder. Assuming the bullet undergoes uniform deceleration and stops in 6.0 cm, find: a) the time taken to stop, b) the impulse on the shoulder, and c) the average force experienced by the man.

38. Consider a 5 gm bullet at a speed of 150 m/sec in the positive x-direction that then hits a 2 kg target which was initially at rest. What is the speed of the target after it is hit by the bullet a) if the bullet is steel and becomes inbedded in the target? b) if the bullet is rubber and bounces off at 180 degrees from the initial path (assume elastic collision) ? c) what is the final speed of the bullet in part-a and part-b ?

*39. (Extra credit problem - you should be able to do this, but this will not be asked on the test) Consider a fly-by of Jupiter and Saturn by a spaceship. On flying by Jupiter the spaceship is deflected by Jupiter's gravity. This deflection can be treated as a glancing elastic collision between Jupiter and the spaceship without worrying about the details of theactual path of the fly-by (we assume the spaceship does not go so low as to enter Jupiter's atmosphere which will generate heat and hence will not be elastic). (a) Suppose Jupiter is in such a position that the spaceship approaches Jupiter with a speed of 40.0 km/sec at 120° and is deflected in such a way that the y-component of the spaceship's velocity is unchanged. Jupiter is going at 13.1 km/sec at 180° (in its orbit around the sun). What is the final speed and direction of the spaceship after the collision if we neglect the gravity due to the sun? Assume the spaceship has a mass of 100 kg and the mass of Jupiter is 1.9 x 1027 kg. HINT: You only need worry about the x-direction. (b) Suppose the spaceship instead approaches Jupiter at 60° and again is deflected in such a way that the y-component of the spaceship's velocity is unchanged. What is the final speed and direction of the spaceship after the collision again neglecting the gravity due to the sun?

40. An astronaut during a spaceship flight had his safety line cut by a sharp edge and finds himself floating beside the ship. a) Can he "swim" back to the spaceship? b) If he has a wrench of 2 kg in his hand and throws this away from the ship at 20 m/sec, can he reach his ship? c) If he is 25 meters from his spaceship, how long will it take him to reach his ship after he throws the wrench assuming the astronaut (plus suit, etc.) has a mass of 70 kg? d) What is the astronaut's weight in pounds?

41. A space shuttle of mass 1000 kg including fuel takes off from a space station by burning its engines. If it wishes to reach a speed of 500 m/sec, how long should it burn its engines if its engines burn one kg of fuel per second and eject it at a speed of 50 km/sec?

Return to top

36 a-1] (30 kg m/s, 0); a-2] (3 m/s, 0); a-3] 45 J; b-1] (30 kg m/s, 0); b-2] (7 m/s, 0); b-3] 165 J; c-1] (30 kg m/s, 0); c-2] (-1 m/s, 0); c-3] -75 J; d-1] (30 kg m/s, 0); d-2] (3 m/s, 4 m/s); d-3] 45 J. The same force and time imply the same change in momentum, but the different initial velocities cause a different distance through which the force acts causing different energy changes.

37. (a) 1.2 millisec; (b) 0.5 kg m/s; (c) 416.67 Nt.

38. (a) 0.374 m/s; (b) 0.748 m/s; (c) 0.374 m/s, 149.25 m/s

39*.(a) v = (-6.2 kg/s, 34.64 km/s) = (35.2 km/s, 100.1°) - ship has slowed down. (b) v = (-46.2 km/s, 34.64 km/s) = (57.7 km/s, 143.1°) - ship has speeded up!

40. (a) no; (b) yes; (c) v = 0.57 m/s, t = 43.75 s; (d) in outer- space his weight would be zero, on the Earth it would be 154 pounds.

41. 9.95 kg or 9.95 sec.

Return to top