# Physics 1303,1403 - Spring 1998

# Homework Assignment #4

**Due:** 5 March

### READING

Chapters 7 and 8.
### QUESTIONS

Chapter 7 - 1, 5, 6, 7, 11, 15.

Chapter 8 - 1, 2, 3, 12.
### PROBLEMS

Chapter 7 - 3, 6, 17, 24, 32, 36, 43, 44, 50, 51, 58.

Chapter 8 - 4, 6, 10, 11, 16, 20A, 20, 25A, 25, 30, 35, 57.
### ANSWERS

These are the ANSWERS only, not the SOLUTIONS. It is not sufficient to
copy these and turn them in as homework. You must show your work.
**Q 7-1)** The displacement ds is perpendicular to the force F.

**Q 7-5)** No.

**Q 7-6)** It quadruples.

**Q 7-7)** The speed is unchanged (the direction of velocity might change).

**Q 7-11)** The more massive bullet has twice the kinetic energy of the
smaller mass bullet.

**Q 7-15)** Pushing on a wall: ds=0. No displacement, no work.

Tension in a string that keeps a particle in uniform
circular motion: F is perdendicualr to ds, no work.

**Q 8-1)** If all the forces on the ball were conservative, the ball
would swing back exactly to the point from which it was released. However,
air resistance is non-conservative and it happens to decrease, not increase,
the total mechanical energy. The ball will swing back to a lower height.

**Q 8-2)** Yes.

**Q 8-3)** They will NOT agree on the VALUE of the potential energy, but
they will agree on the CHANGE in potential energy and on the
kinetic energy.

**Q 8-12)** K is a maximum at launch and just before hitting the ground.
U is a maximum at the highest point in the arc.

**P 7-3)**
- 31.9 J
- 0 J
- 0 J
- 31.9 J

**P 7-6)**
- 294,000 J
- 294,000 J

**P 7-17)**
- 16.0 J
- 36.9
^{o}

**P 7-24)**
- 24.0 J
- -3.00 J
- 21.0 J

**P 7-32)**
- 1.20 J
- 5.00 m/s
- 6.30 J

**P 7-36)**
- 60.0 J
- 60.0 J

**P 7-43)** 1.25 m/s

**P 7-44)**
- 0.791 m/s to the left
- 0.531 m/s to the left

**P 7-50)** 234J

**P 7-51)**
- 63.9 J
- -35.3 J
- -9.5 J
- 19.1 J

**P 7-58)** 830 N

**P 8-4)**
- Ax
^{2}/2 - B^{3}/3
- change in U = 5A/2 - 19B/3

change in K = -5A/2 + 19B/3 = -(change in U)

**P 8-6)**
- 80.0 J
- -80.0 J
- 6.32 m/s

**P 8-10)** 2.94 m/s

**P 8-11)** v = sqrt(3 g R)

N = 0.098 newtons downward

**P 8-16)**
- v = sqrt(v
_{o}^{2} + 2gR + 2gR cos theta)
- sqrt(gR)
- No.

**P 8-20A)**

**P 8-20)** 1.96 m

**P 8-25A)**

**P 8-25)** 3.74 m/s

**P 8-30)** 0.344 m

**P 8-35)** 0.327

**P 8-57)**

Please report any corrections to
Professor Scalise.

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