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Unformatted text preview: Homework 1
Solution 1. The time Trequired for one complete oscillation of a mass m on a spring of force constant k is: T = 2711,13 .Find the dimensions It must have for this equation to be dimensionally correct. a. [Mpm
b. [LyLT]
c. {M' 1H
d. [M]![T2] We can express the constant It in terms of the other variables: T=2Jr 5
k
T2: 2 2 1’1
mm
m
k=(2ﬂ')2F The only dimensionﬁil quantiﬁes are m and T, having dimensions [M] (mass) and [T] (time)
resPectively. So the constant It must have dimensions [M] f[T]2. 2. The ﬁrst several digits of it are known to be ‘1 =3141592653 589?9.... What is 11 10 ﬁve
signiﬁcant ﬁgures?
Answer (nuts: be exact to be correct. since 3.14159 would be to 6 signiﬁcant ﬁgures) 3. Nerve impulses in giant axons of the squid can travel with a speed of 20.0 ma's. How fast is this in (a) {his and (b) mifh?
a. 20mm * (Immune m)= b. 20 info * (1 mifl509 m) * {3500 5” hr) = m 4. The golfer in the ﬁgure sinks the hall in two putts, as shown. What is (a) the distance traveled by the ball. and (b) the displacement of the ball? First putt —~—lﬂm a. Distance = lﬂm+ 2.5 m+ 2.5 m=15 m
b. Displacement = xr— x0 = lﬂm — 0111 = 10 In
5. You jog at 3.0 mifh for 4.0 mi, then you jump into a car and drive for another 6.0 mi. With what
average speed must you drive if your average speed for the entire 9.!) miles is to be 4.4 M? a. The average speed is 5.“. = (total distanceMtotal time}. We know that the average speed
must be 4.4 mifh and we know the total distance is 9.0 miles. So the only thing we need
to find is the total time. Since the motion is done in two parts, we can ﬁnd how long it
takes for the jogging part, and then ﬁnd how long it takes for the driving part. For the jogging part, we have constant velocity motion with velocity +3.0 mifh for 4.0
mi. That means the time for the 1" part is: ti = 4.01111 213.0 mifh} = 1.333 hrs. For the driving part, the corresponding time will depend on how fast we drive. That is
what they are asking us, the speed which we need to drive so get the given average speed,
call the driving Speed 54 So the time for the driving part is: t; = 6.0 mi 2’ (54 mifh]. So the total time is t] T :2, such that the average speed is 4.4 mi/h, in other words 4.4 . = 9.01111 = 9.01111 .
1.333 h+t2 1333h+ 6.01m
. 5d
So the rest is just an algebraic problem, we need to solve for 5‘.
1333 h+ 6.01m = 9.0 on
s:1 4.4 111th
6.0 mi: 9.0 mi 4333 h
s‘i 4.4 tnn'h
sd = l
6.0 mi ﬂ 4333 h
4.4 mlfh
5d : 9 0 6.0 on = 6.0 m1 = 6.0 rm 2 8.42 mit'h
4.4 mtfh 6. Air bags are designed to deploy in 1:10 ms. Assume that the from surface of the bag starts from
rest and expands 0.30 m with constant acceleration. Estimate the acceleration of the front surface
of the bag as it expands. Express your answer in terms of the acceleration of gravity 3. a. We need to ﬁnd the acceleration of the from surface. We are told to assume constant acceleration. So we can use any of the equations that deal with constant acceleration. We do not know the final velocity, so that only leaves us with: x=xo+vot+ Vzatz if we chose our coordinates such that at t=0 the bag will just start expanding from rest
and we set the origin of our xaxis to coincide with the initial position of the bag, we
can simplify the equation. With that choice of coordinates, x0=0 and v0 = 0. So we
are leﬁ with: a: 5’; at2 Now we solve for a (the time is given in ms, and lms is 10'3 s): a=2xft2 = 2(030 mm 10 x 10'3 sf = 6000 mtg.2 Since they ask us to express this acceleration in terms of the acceleration of gravity, and
we know that g=9.81 mlsz, this acceleration is equal to:
a: 6000 mist (1 goat mfszj =
T. On October 9, 1992, a 2?—pound meteorite struck a car in Peekskill, NY, leaving a dent x=25 cm
deep in the trunk. If the meteorite struck the car with a speed of v=6ﬂ0 mfs, what was the
magnitude ofits deceleration, assuming it to be constant?
a. The meteorite reaches the car With v=600 mils, let's set this to he i=0, so the initial
velocity is vﬁ=600 rots. Let’s also set this as the origin of our coordinate system, such
that the meteorite is at xo=0 at t=0 and reaches x=25 cm or x=0.25 to when it comes to a
stop. With this choice of axis, the positive x direction is DOWN. Because it comes to a
stop, we know that its ﬁnal velocity will be v20. So we can use the equation
v2 = v02 + 2a(xxo}
For this case, we then have:
0 = {600 ms) 1 + 2a{0.25 me}
So we just need to solve for the value of a:
4000 nuts)2 = 2a(0.25 m)
a = 4600 me)2 I (2 * 025 m)
a = 4360000 mgfszli’OjO m =
Note that the acceleration is negative, because the acceleration is pointing UP, Le, it is decelerating the meteorite which has a velocity pointing DOWN. The problem asks for the magnitude of the deceleration, so this is just a = 720000 mfsz. 3. A bicyclist is ﬁnishing his repair of a flat tire when a friend rides by at 5 refs. Two seconds later, the bicyclist hops on his bike and accelerates at 3 W52 until he catches his friend. From the
moment he starts accelerating, how much time does it take until he catches his ﬁiend? :1. Let’s set the time where the bicyclist starts accelerating as the t=0. Let‘s also set the
initial position of the bicyclist as torO. He undergoes motion with constant acceleration.
starting from rest, so his equation of motion, it his position as a function of time, will
be:
it,,=it0+w2lt+Vast2
where xb denotes the position of the bicyclist. Because of our choice of axis, x0=0 and
“3:0 so this simpliﬁes to:
xh = is at2= 1/: {3 nifsz} t2
His friend undergoes motion with constant velocity. We chose t=0 at the moment when
the bicyclist starts accelerating. At that moment, his friend is two seconds ahead. Since
the friend is moving at 5 rots, he is 5 this * 2 s = it] m in front of the bicyclist. This can
be expressed by the constant velocity equation of motion for the friend:
to: v{t+2s) = 5 nifs (t + ls] = (5 nits] 't + 10 in
The question asks us for the time from the moment the bicyclist starts accelerating until
he catches his friend. When does he catch his friend? When they are both at the same
position! This means that to = xi, is the condition of “catching his friend“. We use the equations to solve for the time that satisﬁes that condition. Then the answer
is the time difference between the moment he catches his friend (the time that solves the
equation xeb) and the moment he started acceleration, which for our choice of coordinates was tee. So all we need now is to ﬁnd the time that solves the equations for arm. This is now just an algebra problem of solving for t in the equation: Ib=1f 1 1
—: = 1+2
2:1 WC ) l—3:1=5:+1t)
2 1.5rz—Sr—10= 0
We use the general solution to the quadratic equation
(See Appendix A if yOu need a refresher} _ 2—
=bi—r—— Myriam a = 15,45 = —5,c = 10
a r— —(—5) i ,ltsf — 4(1 .5}(—l G) 2(1.5) 5i¢25+6ﬂ 3 5 iJSS
3 5:53.22 3
one solution is negative, I f: we need the positive time solution, so IE=14.22 14474 5 Seagulls are often observed dropping clams and other shellﬁsh Front 3 height to the rocks below,
as a means of opening the shells. If a seagull drops a shell from rest at a height of h=20 in, how
fast is the shell moving when it hits the rocks in MS? a. The shell is in free fall1 constant acceleration of g=9.81 mfsz. We need to ﬁnd the ﬁnal
velocity. The shell starts from rest, so v0=0. if we choose our coordinates such that the
initial position is x0=0 and the ﬁnal position is x=20 In, we can use the equation: 1;2 = no2 + 2a(x x0) v2 2 0+ 209.81 nuszxzoo 111—0)
2 = 392.4 mite? v = 4392.4 nﬂs2 =19.3 mis If}. Wrongly called for a foul, an angry basketball player throws the ball straight down to the floor. If the ball bounces straight up and returns to the ﬂoor 3 s after ﬁrst striking it, what was the ball's greatest height above the ﬂoor? a. The ball is undergoing ﬁ'ee—fall from the moment it leaves the ﬂoor. Unfortunately, we
don‘t know the velocity with which it leaves the floor going up. But we do know that it
starts from the oor and ends back in the ﬂoor in 3 s. From the symmetry of the motion.
we know that it must be at its maximum height at exactly 3t2 s = 1.5 s. We also know,
that at this moment. its instantaneous velocity is exactly zero. So if we study the motion
from the time of max height until the moment it reaches the floor, its displacement will
be the maximum height, the initial velocity will be equal to ID and the acceleration will be the constant g=9,31 1an2 for all objects in free—fall, so we can calculate use: xm = x0+v01 +égt‘2 xm = o + o +%(9.31 moon .5 s}2 xm =%(9.81mﬁsz)(2.25 52') = 1.04 ml ...
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 Fall '09
 WOODAHL
 Physics

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