PHYSICS 121 FALL 2003 - Homework #1 - Solutions

Problems from Chapter 1: 3E,7P,21P

3. Using the given conversion factors, we find

(a) the distance d in rods to be

d = 4.0 furlongs =(4.0 furlongs)(201.168m/furlong)

5.0292m/rod= 160 rods ,

(b) and that distance in chains to be

d =(4.0 furlongs)(201.168m/furlong)

20.117m/chain= 40 chains .

7. The volume of ice is given by the product of the semicircular surface area and the thickness. Thesemicircle area is A = πr2/2, where r is the radius. Therefore, the volume is

V =π

2r2 z

where z is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have

r = (2000 km)(

103 m1km

)(102 cm1m

)= 2000 × 105 cm .

In these units, the thickness becomes

z = (3000m)(

102 cm1m

)= 3000 × 102 cm .

Therefore,V =

π

2(2000 × 105 cm

)2 (3000 × 102 cm

)= 1.9 × 1022 cm3 .

21. We introduce the notion of density (which the students have probably seen in other courses):

ρ =m

V

and convert to SI units: 1 g = 1 × 10−3 kg.

(a) For volume conversion, we find 1 cm3 = (1 × 10−2 m)3 = 1 × 10−6 m3. Thus, the density in kg/m3

is

1 g/cm3 =(

1 gcm3

)(10−3 kg

g

)(cm3

10−6 m3

)= 1 × 103 kg/m3

.

Thus, the mass of a cubic meter of water is 1000 kg.

(b) We divide the mass of the water by the time taken to drain it. The mass is found from M = ρV(the product of the volume of water and its density):

M = (5700m3)(1 × 103 kg/m3) = 5.70 × 106 kg .

The time is t = (10 h)(3600 s/h) = 3.6 × 104 s, so the mass flow rate R is

R =M

t=

5.70 × 106 kg3.6 × 104 s

= 158 kg/s .

PHYSICS 121 FALL 2003 - Homework #1 - Solutions

Problems from Chapter 2: 1E,5P a)&b),13P,17E,23E,37P,43E,48P

1. Assuming the horizontal velocity of the ball is constant, the horizontal displacement is

∆x = v∆t

where ∆x is the horizontal distance traveled, ∆t is the time, and v is the (horizontal) velocity. Convertingv to meters per second, we have 160 km/h = 44.4m/s. Thus

∆t =∆x

v=

18.4m44.4m/s

= 0.414 s .

The velocity-unit conversion implemented above can be figured “from basics” (1000 m = 1 km, 3600 s= 1 h) or found in Appendix D.

5. (a) Denoting the travel time and distance from San Antonio to Houston as T and D, respectively, theaverage speed is

savg 1 =D

T=

(55 km/h)T2 + (90 km/h)T

2

T= 72.5 km/h

which should be rounded to 73 km/h.

(b) Using the fact that time=distance/speed while the speed is constant, we find

savg 2 =D

T=

DD/2

55 km/h + D/290 km/h

= 68.3 km/h

which should be rounded to 68 km/h.

13. Since v = dxdt (Eq. 2-4), then ∆x =

∫v dt, which corresponds to the area under the v vs t graph. Dividing

the total area A into rectangular (base×height) and triangular (12base×height) areas, we have

A = A0<t<2 + A2<t<10 + A10<t<12 + A12<t<16

=12

(2)(8) + (8)(8) +(

(2)(4) +12

(2)(4))

+ (4)(4)

with SI units understood. In this way, we obtain ∆x = 100 m.

17. We represent its initial direction of motion as the +x direction, so that v0 = +18 m/s and v = −30 m/s(when t = 2.4 s). Using Eq. 2-7 (or Eq. 2-11, suitably interpreted) we find

aavg =(−30)− (+18)

2.4= −20 m/s2

which indicates that the average acceleration has magnitude 20 m/s2 and is in the opposite direction tothe particle’s initial velocity.

23. The constant-acceleration condition permits the use of Table 2-1.

(a) Setting v = 0 and x0 = 0 in v2 = v20 + 2a(x − x0), we find

x = −12

v20

a= −1

2

(5.00× 106

−1.25× 1014

)= 0.100 m .

Since the muon is slowing, the initial velocity and the acceleration must have opposite signs.

(b) Below are the time-plots of the position x and velocity v of the muon from the moment it entersthe field to the time it stops. The computation in part (a) made no reference to t, so that otherequations from Table 2-1 (such as v = v0 + at and x = v0t+ 1

2at2) are used in making these plots.

...............................................................................................................................................................................................................................................................................................................................................................................................

.....................................

.............................................................

10 20 30 40

t (ns)

0

2.5

5.0

7.5

10

x (cm)

................................................................................................................................................................................................................................................................................................................................................................................................

10 20 30 40

t (ns)

0

2.0

4.0

6.0

8.0v (Mm/s)

37. We denote tr as the reaction time and tb as the braking time. The motion during tr is of the constant-velocity (call it v0) type. Then the position of the car is given by

x = v0tr + v0tb +12at2b

where v0 is the initial velocity and a is the acceleration (which we expect to be negative-valued since weare taking the velocity in the positive direction and we know the car is decelerating). After the brakesare applied the velocity of the car is given by v = v0 +atb. Using this equation, with v = 0, we eliminatetb from the first equation and obtain

x = v0tr − v20

a+

12

v20

a= v0tr − 1

2v20

a.

We write this equation for each of the initial velocities:

x1 = v01tr − 12

v201

a

and

x2 = v02tr − 12

v202

a.

Solving these equations simultaneously for tr and a we get

tr =v202x1 − v2

01x2

v01v02(v02 − v01)

and

a = −12

v02v201 − v01v

202

v02x1 − v01x2.

Substituting x1 = 56.7m, v01 = 80.5 km/h = 22.4m/s, x2 = 24.4 m and v02 = 48.3 km/h = 13.4 m/s,we find

tr =13.42(56.7) − 22.42(24.4)(22.4)(13.4)(13.4 − 22.4)

= 0.74 s

and

a = −12

(13.4)22.42 − (22.4)13.42

(13.4)(56.7) − (22.4)(24.4)= −6.2 m/s2 .

The magnitude of the deceleration is therefore 6.2 m/s2. Although rounded off values are displayedin the above substitutions, what we have input into our calculators are the “exact” values (such asv02 = 161

12 m/s).

43. We neglect air resistance for the duration of the motion (between “launching” and “landing”), so a =−g = −9.8 m/s2 (we take downward to be the −y direction). We use the equations in Table 2-1 (with∆y replacing ∆x) because this is a =constant motion.

(a) At the highest point the velocity of the ball vanishes. Taking y0 = 0, we set v = 0 in v2 = v20 −2gy,

and solve for the initial velocity: v0 =√2gy. Since y = 50m we find v0 = 31 m/s.

(b) It will be in the air from the time it leaves the ground until the time it returns to the ground(y = 0). Applying Eq. 2-15 to the entire motion (the rise and the fall, of total time t > 0) we have

y = v0t − 12gt2 =⇒ t =

2v0

g

which (using our result from part (a)) produces t = 6.4 s. It is possible to obtain this withoutusing part (a)’s result; one can find the time just for the rise (from ground to highest point) fromEq. 2-16 and then double it.

(c) SI units are understood in the x and v graphs shown. In the interest of saving space, we do notshow the graph of a, which is a horizontal line at −9.8m/s2.

...........................................................................................................................................................................................................................................

.................................................................................................................................................................................................................................................................................

2 4 6 8 t0

20

40

60 y ............................................................................................................................................................................................................................................................................................................................................................................

2 4 6 8t

0

−20

20

40v

48. We neglect air resistance, which justifies setting a = −g = −9.8 m/s2 (taking down as the −y direction)for the duration of the motion. We are allowed to use Table 2-1 (with ∆y replacing ∆x) because this isconstant acceleration motion. The ground level is taken to correspond to the origin of the y axis. Thetotal time of fall can be computed from Eq. 2-15 (using the quadratic formula).

∆y = v0t − 12gt2 =⇒ t =

v0 +√

v20 − 2g∆y

g

with the positive root chosen. With y = 0, v0 = 0 and y0 = h = 60 m, we obtain

t =√2gh

g=

√2hg

= 3.5 s .

Thus, “1.2 s earlier” means we are examining where the rock is at t = 2.3 s:

y − h = v0(2.3)− 12g(2.3)2 =⇒ y = 34 m

where we again use the fact that h = 60 m and v0 = 0.

Additional Problems:

I. If a runner runs at an average speed of 7.0 miles/hour while the runner's heartbeats at 100 beats/minute, how many times does the runner's heat beat during a3.0 mile run?

The time it takes the runner to run 3.0 miles is∆t = D/savg = 3.0 mile/(7.0 mile/hour)

= 0.4286 hour x (60 min/hour) = 25.71 minThe number of heartbeats in 25.71 min is

#of beats = (100beats/min)*25.71 min = 2571 beats = 2600 beats rounded to 2 sig. figs.

II. A mouse moving at a constant velocity of 3.0 m/s passes a cat at rest. The catimmediately begins chasing the mouse moving with constant acceleration. If thecat catches the mouse a distance of 4.0 m from its starting point,a. How long (how much time) does it take for the cat to catch the mouse?

The time it takes for the mouse (and the cat) to move 4.0 m is∆t = ∆x/vavg = 4.0m/(3.0m/s) = 1.333 s = 1.3 s rounded to 2 sig figs

b. What is the cat's acceleration?

The cat is moving with constant acceleration so the cat's position is given by: x = xo + vot + (1/2)at2

you can assume that xo = 0. The cat is starting from rest so, vo = 0 so: x = (1/2)at2

Solving for a: a = 2x/t2 We already know that when x = 4.0m t = 1.333s so:

a = 2(4.0m)/(1.333s)2 = 4.502 m/s2 = 4.5 m/s2 rounded to 2 sig figs

c. What is the average speed of the cat while it is chasing the mouse?

The cat and mouse undergo the same displacement in the same time. So theaverage velocity of the cat must be the same as the velocity of the mouse:

savg = 3.0 m/s

III. In problem 7P in Chapter 1, you found that the volume of ice in Antarctica is 1.9 x 1022 cm3. If all of the ice in Antarctica were to melt, how many meterswould sea level rise. You will need the following information: The density of ice is920 kg/m3 and the density of liquid water is 1000 kg/m3. The total surface area ofthe Earth's oceans is 3.60 x 108 km2.

The mass of the ice is: M = ice density x ice volumeThe mass of the melted water is the same as that of the ice:

M = (water density)(water volume)The volume of the water is given by:

water volume = (area of oceans)(rise in sea level)Solving for rise in sea level:

rise in sea level = (water volume)/(area of oceans) = (M/(water density))/(area of oceans)

= ((ice density)/(water density))(ice volume)/(area of oceans)= (920 kg/m3/1000 kg/m3)(1.9x1022 cm3)/(3.60 x 108 km2)= 48.56 m= 49 m rounded to 2 sig figs