Practice problems

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Romeo sets off in his spaceship to marry Juliet, who lives on the planet Lovelon, 10 light-years away from earth. He can accelerate at 15 m/s2, and must decelerate at the same rate, in order to stop at Lovelon. Unknown to him, Juliet has taken an earlier shuttle back to Earth and she will wait there. Since there is no communication, Romeo will have to travel back again! Romeo left yesterday, on April 30, 1997. What is the earliest year in which they can be wed? Romeo and Juliet are both classical persons, and they ignore any relativistic effects! 1 light-year is about 9.46 x 1015 m.

(A) 1999 (B) 2004 (C) 2007 (D) 2047 (E) they will die from old age first

This is a straightforward s = (1/2)a t2 problem, to ease students into the exam. It's easy to calculate the time to get to Lovelon, and each light year represents

1 light-year = 365 days/year x 24 hours/day x 3600 seconds/hour x 3 x 108 m/s = 9.46 x 1015 m.

If Romeo travels all the way there at 15 m/s2, then 10 light-years will take him a time given by 10 x 9.46 x 1015 = 1/2 x 15 (time)2 giving 1.123 x 108 seconds, or 3.56 years, and he will be moving at 16.85 x 108 m/s, which in Startrek language is Warp 5.6! This won't do him much good of course, because he will pass Lovelon at this speed. Romeo has to travel half-way to Lovelon, 5 light-years, then reverse his rockets to decelerate for another 5 light years at 15 m/s2, so that he arrives at low speed for docking. This splits the journey exactly in half, due to the obvious symmetry of the motion. At the half-way point he'll be moving at 11.9 x 108 m/s, which is near enough Warp 4!

It'll take him 7.94 x 107 sec to get half-way, about 2.51 years, so the total time to get to Lovelon and then back to earth will be 4 times this, or 10.07 years. They can wed in 2007 (C).

illustration - acrobat performing a hand stand with one foot resting against a vertical wall

An overweight acrobat, "weighing" in at 115 kg, wants to perform a single hand stand. He tries to cheat by resting one foot against a smooth frictionless vertical wall. The horizontal force there is 130 N. What is the magnitude of the force exerted by the floor on his hand?
Answer in N

(A) 1134 (B) 1257 (C) 997 (D) 1119 (E) 1127

The acrobat has a force acting on his hand that we resolve into two perpendicular components: the vertical one is the reaction to the weight (115 x 9.8 N = 1127 N) and the horizontal one balances the 130 N force from the wall. These two forces give a resultant force F of
F = √(11272 + 1302) = 1134 N (A).

For each breath that you take, how many of the air molecules would also have been breathed by the patron saint of Physics, Sir Isaac Newton during his lifetime (1642-1727)?
The atmosphere is about 8 km high, and the molecules in the air each occupy a space representing a little cubic box about 33.3 x 10-9m along a side. The earth's radius is 6.37 x 106 m.
Make reasonable assumptions for any data that you need!

(A) 6 (B) 6 x 10³ (C) 6 x 106 (D) 6 x 109 (E) 6 x 1012

If each breath of Sir Isaac's (and ours) is about 1 litre = 10-3 m3, and they are 3 seconds apart, then in 1727-1642=85 years, he will have had 85 x 365 x 24 x 3600 /3 = 8.93 x 108 breaths, for a total volume of 8.93 x 105 m3. We will assume that the air mixes well enough that we do not have to worry about air being breathed twice.
The total volume of the atmosphere is 4 RE 2 h, where RE is the Earth's radius, and h is the height of the atmosphere, giving 4.08 x 1018 m3. The fraction of air molecules ever breathed by the patron saint of Physics is thus 8.93 x 105 / 4.08 x 1018 = 2.19 x 10-13.
The number of molecules in each breath of ours is the density 1/(3.3 x 10-9)3 = 2.78 x 1025 m-3, multiplied by the volume of each breath, 10-3 m3, or 2.78 x 1022 molecules. Multiplying by the fraction breathed by Newton, each breath of ours has about 6.08 x 109 molecules also breathed by him (D). Since we have about 9 x 108 breaths, each breath of ours has about 7 molecules also breathed by Isaac Newton.

The ice storm this winter in the province of Quebec strained many wires to the breaking point. In a particular situation, the transmission pylons are separated by 500 m of wire. The top grounding wire is 15° from the horizontal at the pylons, as in the diagram, and has a diameter of 1.5 cm. The steel wire has a density of 7860 kg/m3. When ice (density 900 kg/m3) built up on the wire to a total diameter of 10.0 cm, the wire snapped. What was the breaking stress (force/unit area) in N/m2 in the wire at the breaking point?
You may assume the ice has no strength.

(A) 7.4 x 107 (B) 4.4 x 108 (C) 2.6 x 106 (D) 1.15 x 107 (E) 6.7 x 108

A free-body diagram shows us that the vertical component of the tension in the two ends of the wire, 2Tsin(15°), must equal the total loaded weight W.
The volume of the wire is \(\pi R_{w}2L\), where \(R_{w}\) is the radius of the wire, and the total volume of ice is \(\pi(R_{i}^{2}-R_{w}^{2})L\), where \(R_{i}\) is the radius of the ice-covered wire. Multiplying by the respective densities, the total mass of the 500m-long ice-covered wire is 694.5 + 3531.5 = 4226 kg, having a total weight of 41415 N. The tension in the wire is thus \(\frac{41425}{2sin15^{\circ}} = 80000 N\), and dividing by the cross-sectional area of 1.77 x 10-4 m3, the stress is 4.52 x 108 \(\frac{N}{m^{2}}\) (B).

Romeo is landing back on Earth to claim his bride, Juliet. His spaceship is descending vertically at 50 m/s and he is 500m up when he throws a bouquet of Lovelon flowers horizontally (from his point of view) out of his ship to land near Juliet, who is waiting at the dockport, 40 m from his landing spot. How fast must Romeo throw his flowers in m/s so that they land at Juliet's feet? (Lovelon flowers are very slippery, and do not experience air resistance. They can also absorb quite a shock on landing and still look fresh!).

(A) 1.96 (B) 2.52 (C) 4.00 (D) 5.24 (E) 6.44

Romeo's flowers are 500 m up, have a downward velocity component of 50 m/s, and they will
strike the surface of the ground in a time given by \(500 = 50t + \frac{9.8t^{2}}{2}\), which solves to
give t = 6.21 seconds. The flowers must cover 40 m horizontally in this interval, so Romeo
must throw them out sideways at \(\frac{40}{6.21} = 6.44 \frac{m}{s}\) (E).

A teenager is downloading pirated, compressed songs in MP3 format from the Internet. Each song is about 3 minutes long when played, and needs about 2.9 megabytes of computer storage. To fill a CD requires about 70 minutes of music. If a 56K modem is being used which gives on average about 30 kilobits/sec of data flow, how many minutes does it take to download a full CD's worth of songs, ignoring setup times and filename entries? A byte represents 8 bits.

(A) 301 (B) 203 (C) 187 (D) 38 (E) 14

With each song being 3 minutes long, there are 23.33 songs on a 70-minute CD. This requires a total storage of 2.9 megabytes x 23.33 = 67.67 megabytes. The modem processes about 30 kilobits/sec, which is 3.75 kilobytes/sec. At this rate, 67.67 megabytes will take (67.67x106)/(3750) = 18045 seconds or 301 minutes (A) or 5.01 hours! 

"A Decade of SIN Plus Sixteen" is a compilation of past exams and solutions, from 1969 - 1994.
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