Physics 101
Lawrence Weinstein
Homework set 2
Solutions

1-Steps:
-------------------
Chapter 3:
4: Speed = distance / time = 24 m / 0.5 s = 48 m/s

12: acceleration = change in velocity / time
a = delta v / t = 25 m/s / 5 s = 5 m/s^2

Chapter 4:
4: An apple weighs 1 N.  Using F = mg, we have m = F/g
m = 1 N / 10 m/s^2 = 0.1 kg  (since 1 N = 1 kg m / s^2)

Weight (lb) = Weight (N) * ( 1 lb / 4.448 N )
Weight = 1 N = 1 N * (1 lb / 4.448 N) = 0.22 lb 

8:  a) F = ma -> a = F/m = 20 N / 2 kg = 10 m/s^2
b) Now F_net = 20 N - 4 N = 16 N
a = 16 N / 2 kg = 8 m/s^2


Exercises:
--------------------------
Chapter 3:
4: The speeds are the same (300 km/h) but the velocities are different
because the directions are different.

16: a) Anything that changes direction without changing speed undergoes
acceleration.  This would include the Moon (or any other satellite), a
car travelling around a curve at constant speed, a ball on a string
whirled around in a circle.

b) Since acceleration = change in velocity / time, if the velocity is
constant, then the _change_ in velocity = delta v = 0.  Therefore, if
the velocity is constant, then the acceleration must be zero.

18: If you are driving your car at a certain speed in a straight line,
then there are four ways your velocity can change.  You can a) speed
up, b) slow down, c) turn right or d) turn left.  If you speed up,
then your acceleration is in the same direction as your velocity.  If
you turn, then your acceleration is in a different direction than your
velocity (eg: to the left or right).  If you slow down, then the
acceleration is in the opposite direction to the velocity.

(Note that this answer is much longer than your answer needs to be.
The last sentence is all you need.)

34: Use the equation for distance travelled at constant acceleration:
d = (1/2) a t^2 (see p49).
For ten 'counts':
d1 = (1/2) a (10)^2 = (1/2) a 100
For twenty counts:
d2 = (1/2) a (20)^2 = (1/2) a 400
Thus, the second distance is four times larger than the first so that 
d2 = 4 * 1 cubit = 4 cubits.

We'll be looking at lots of questions of this type (ie: if you double
this quantity, what happens to that one).


Chapter 4:

6: Apples have the same mass whether they are on the Earth or the
Moon.  However, on the Moon, the force of gravity is less so that they
weigh less.  Pounds are a unit of weight, kilograms are a unit of
mass.  Therefore, an amount of apples that weigh one pound on
the Earth would weigh a lot less on the Moon.  Therefore, one pound of
apples on the Moon contains more apples.

However, gravity does not change the amount of mass so that 1-kg of
apples is the amount on the Earth and the Moon.

10: When your hand hits the wall, it accelerates until its velocity is
zero (until it stops).  The wall needs to exert more force on your
hand to provide that acceleration (ie: to stop it) if you're carrying
a heavy load.  This is Newton's 2nd Law, F = ma.

26: Newton's 1st Law says that motion with constant velocity requires
no NET (total) force.  There are other forces on the cart in addition
to the force you are applying.  Your force plus the force of friction
totals to zero so that the NET force is zero.

28: a) At terminal velocity, the skydiver is not accelerating.
Therefore, the net force is zero.  Therefore the force of air
resistance is equal to and opposite the force of gravity.  Therefore,
the skydiver is not in freefall since gravity is not the only force.

b) the satellite above the atmosphere is in free fall since a) gravity
still pulls on it and b) there is no air resistance since it is above
the atmosphere.

--------------------
Problems:

Chapter 3: 
4: I will define + as upward and - as downwards
a) at the highest point, v = 0
b) 1 s before then, v = 10 m/s
c) delta v = v_after - v_before = 0 m/s - 10 m/s = -10 m/s
d) 1 s after v = -10 m/s
e) delta v = -10 m/s - 0 m/s = -10 m/s
f) from 1 s before to 1 s after
delta v = -10 m/s - (10 m/s) = -20 m/s
g) from 1 s before to the highest point: 
a = delta v / t = -10 m/s / 1 s = -10 m/s^2
from the highest point to 1 s after, 
a = delta v / t = -10 m/s / 1 s = -10 m/s^2
from 1 s bfore to 1 s after
a = delta v / t = -20 m/s / 2 s = -10 m/s^2
(You can just write -10 m/s^2 for the answer.  You don't need to do
all three cases.)

8: This will work if you choose any distance.  I will do this for a
specific distance and then for an arbitrary distance (either method is
OK).  I choose to use 120 km for the distance travelled (since it is
divisible by both 40 and 60).

outward travel time = 120 km / 40 km/h = 3 h
return  travel time = 120 km / 60 km/h = 2 h
speed = total distance / total time = 240 km / 5 h = 48 km/h

Here's the general proof:
t_out = d / 40
t_ret = d / 60
t_total = d(1/40 + 1/60) = d(3/120 + 2/120) = d(5/120) = d/24
d_total = d+d = 2d
speed = d_total / t_total = 2d / (d/24) = 48 km/h


9: This is a two step problem.  First we need to figure out the time
to fall, then we can use that to get the speed.

d = (1/2) g t^2
t = sqrt(2d/g) = sqrt(2000 m / 10 m/s^2) = 14.1 s

v = a t = 10 m/s^2 * 14.1 s = 140 m/s

(Since 1 m/s = about 2 mph, this is about 300 mph.  OUCH!)

--------------------------
Estimation:
You grow about 6 feet in 20 years so that v = 6 ft / 20 yr = 0.3 ft/yr
a) convert to inches: v = 0.3 ft/yr * (12 in / 1 ft) = 4 in/yr
b) now convert to m/s
v = 0.3 ft/yr * (1 yr / 3.1*10^7 s) * (0.3 m / 1 ft) = 3*10^(-9) m/s
Wow.  That's slow!
c) now convert to mph:
v = 3*10^(-9) m/s * (2 mph / 1 m/s) = 6*10^(-9) mph.

(Note: you can convert m/s to mph exactly, but 1 m/s = 2 mph is close
enough for estimation.)