SOLUTIONS FOR FIRST HOUR EXAM
1. a) Density is mass/volume.
b) You took the mass of a number of nails by measuring them with a balance, and determined the volume of the nails by measuring the
amount of water they displaced in a graduated cylinder. You determined the density by dividing the mass by the volume.
c) The greatest source of error was uncertainty in volume, since most of you had displacements of only 1 or 2 ml, generating huge
erros in your volume. They way to overcome this is to use many more objects, so they would displace more water, and the uncertainty in
reading the volume change would become a smaller percentage of the actual volume change.
2. a) Atomic number is the number of protons in a nucleus, atomic mass is the sum of protons and neutrons in a nucleus, and an isotope
a nucleus with the same number of protons but different number of neutrons.
b) It is the atomic number, i.e.,. number of protons in a nucleus that determines the type of nucleus.
c) In the original carbon nucleus, there are 6 protons and 14 protons+neutrons, or 8 neutrons. If a neutron turns to a proton,
there are now 7 protons and 7 neutrons. This is a nucleus of N 14.
3. Speed is how much distance is covered in a certain time, velocity is the speed in a specified direction, and acceleration is the
rate of change of velocity.
4. Instantaneous speed is the average speed in a very short time interval. Avg speed = dist/time
5. a) The ball travels one meter in two secs, so avg speed = dist/time = 1m/2s = 0.5 m/s
(final speed - initial speed)/time = (1.0m/s -0 m/s)/2 s = 0.5 m/s/s
b) As you did in class, the final inst. speed is twice the average speed, so is 1.0 m/s
c) Acceleration is:
6. a)Here we know that the acceleration is 10 m/s/s, so rearranging the equation above: a = (final speed- initial speed)/time
a x time = final speed - initial speed
if the initial speed is zero, then:
final speed = a x t = 10 m/s/s x 3 s = 30 m/s
b) If the initial speed is zero, and the final speed is 30 m/s, the average speed is 15 m/s
c) If the average speed is 15 m/s and the ball is in the air for 3 secs, then the total dist traveled is 15 m/s x 3 s or 45 m
Alternately, you could have used the equation:
d = 1/2 a t2 = 1/2 x 10 m/s/s x (3s)2 = 45 m
7. You have a couple of ways you could do this. You could use the fact that the ball accelerates by 0.5 m/s/s to reason that its speed
will increase by 0.5 m/sec for each additional second it is in motion. If we extend its time of travel by two additional seconds, its
speed will increase by an additional 1 m/s, so will have a final intantaneous speed of 2 m/s. This implies that its average speed would
be 1.0 m/s. If an object travels at an average speed of 1.0 m/s for 4 secs, it covers 4 meters.
Alternately, you could use the equation in question six above as: dist = 1/2 at2 = 1/2 (0.5 m/s/s)(4 s)2 = 1/2x1/2x16 m = 4 m
8. Mass is a scalar and is the amount of "stuff" or matter in an object. Mass is a measure of inertia, an object's resistance to
changes in motion. Weight is a vector and is the force of gravity acting on an object.
9. See solutions to homework problems.
10. a) For this we use the Pythagorean Theorem to calculate the resultant speed. The result of this calculation is 5 mi/hr.
b) The only component of the swimmer's motion that carries him/her across the river is the swimmer's speed across the river; the river current, since it is parallel to
the banks, has no influence on this motion, so it takes 2 hours to travel 6 miles at 3 mi/hr.
c) The only component of the swimmer's motion that carries him/her downstream is the current of the river. If the swimmer is in the water for 2 hours, the current
will carry the swimmer 8 miles downstrean (2 hours x 4 mi/hr). Notice that the total distance covered by the swimmer in the diagonal path across the river is 10 miles. If
we use the total speed calculated in part a, we are not surprised to find that the swimmer will cover a total distance of 10 miles travelling at a total speed of 5 i/hr for
David B. Slavsky
Loyola University Chicago
Cudahy Science Hall, Rm. 404
1032 W. Sheridan Rd.,
Chicago, IL 60660
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