PHY2060 Test I, Sept. 27, 2011
1. An airplane pilot, in order to fly due east over the ground from his point of origin, has to
fly 15 o North of East in the air due to a cross wind (see sketch). If the plane’s speed in the
air is 135 m/s (~ 300 miles/hour) and if the plane take 2 hours 45 minutes to fly 1440 km
over the ground, what was the speed and direction (draw a diagram making this clear!) of
the cross wind? (units of m/s, 3 sig figs.)
Soln.: a non-right angle triangle has as its base, which is from due west to due east, the
resultant velocity over the ground. A side of the triangle that is 15 o above this base has as
its length 135 m/s. The third side of the triangle, opposite the 15 o angle, is the velocity of
the cross (and following) wind. See the diagram. Using the law of cosines
c2 = a2 + b2 -2a*b*cos(angle c) where angle c is 15 o, c is the length of the cross wind side,
and sides a and b are the 135 m/s velocity vector in the air and the due eastward net
velocity of the plane over the ground. The due eastward net velocity of the plane over the
ground is 1440,000 m/2 hr 45 min = 145.45 m/s. Solving for c gives 38.1 m/s. Define  as
the angle between the wind and the due east velocity of the plane over the ground, and
sin/side opposite  = sin15o/side opposite 15 o
or sin/135 m/s = sin 15o/wind velocity=38.1 m/s implying that =66.5 o.
2. An airplane pilot flying at v=200 m/s at an altitude of 1000 m wishes to release a bomb
in such a way that the plane can have lots of time to fly away from the explosion.
a. If all the pilot did was to launch the bomb in the horizontal direction of the
plane’s flight path, how far horizontally from the launch point (see sketch) would the bomb
impact the ground. (in m, 3 sig figs)
b. Suppose the pilot sets the plane into a 45 o above the horizontal upward climb
just before the bomb releases (the speed of the climbing plane is still 200 m/s and assume
the release point is still at 1000 m.) How far horizontally from the launch point does the
bomb impact the ground? (units of m, 3 sig figs)
a.) y-y0 = -1000 m = ½ (-9.8 m/s2)t2 ; t=14.29 s, x-x0 = v0x*t = 200 m/s * 14.29 s = 2857 m
b.) y-y0 = -1000 m =v0yt + ½ (-9.8 m/s2)t2 ; v0y=200 m/s * sin45o = 141.4 m/s
so we have 4.9 t2 – 141.4 t -1000 = 0 => t= (141.4  sqrt (141.42 – 4*4.9*(-1000)))/9.8
t=(141.4  sqrt (20,000 + 19,600))/9.8 = (141.4  199)/9.8 choose the positive root since the
negative time refers to backtracking the path to get
t=340.4/9.8 = 34.73 s so x-x0 = v0x * t and v0x=cos45 * 200 = 141.4 m/s so x-x0=4911 m or
over 70% further than by method a.)
3. An entrepreneurial PHY2060 student who wants to do more with what they’re learning
proposes a new exciting act for the ice capades – having a clown in a sled on the (assumed)
frictionless ice and firing 2 kg cannon balls in one direction to achieve recoil velocity in the
opposite direction. If the clown + cannon + sled mass = 100 kg plus 4 kg for the 2-shot
ammunition sequence (ignore mass of the gun powder), and if the ejected cannon ball
velocity is 100 m/s relative to the sled, what is the final velocity in m/s (3 sig figs) of the
clown + cannon + sled after 3 shots are fired one after the other? (Note that if the cannon
balls were somehow claimed to be going at 100 m/s relative to the ice rink for both shots,
then more gunpowder would be needed since the second shot comes from a more recoiling
sled. Just conserve momentum in the ice rink and you won't get confused.)
Step 1: pfinal=pinitial = 0, pfinal = 102 kg (1 ball still in sled)*(-)vrecoil in one direction added to
2 kg *(+) (100 m/s - vrecoil) momentum of the first cannon ball in the opposite direction, =>
(102+2) vrecoil step 1 = 200 kg m/s => vrecoil step 1=1.923 m/s (vrecoil is taken as positive and the
sign for the recoil is stated explicitly)
Step 2: same thing over in the rest frame of the ice rink:
pinitial = - 102 * 1.923 = pfinal =
100 kg (no ball left in sled) * (-)vrecoil step 2 + 2kg*(100 m/s - vrecoil step 2)
-196.146 kg m/s = 102 kg * (-)vrecoil step 2 + 200 kg m/s
=> vrecoil step 2 = 3.884 m/s
4. Consider a 750 g baseball bat (around 27 oz, a not unusual baseball bat size) moving
forwards (say along the x-axis) in a plane horizontal to the ground at 18 m/s at the time it
contacts an incoming 140 g (5 oz) fastball moving at -40 m/s along the x-axis in the same
horizontal plane (i. e. a one dimensional ‘head on’ collision.) If the ball after being struck
is traveling along the +x axis at 37 m/s, what is the bat speed and direction after the
impact? (2 sig figs)
Pinitial = 0.75 kg * 18 m/s – 0.14 kg * 40 m/s = 7.9 kg m/s
Pfinal = 0.75 kg * ? + 0.14 * 37 m/s, so 0.75 * ? = 7.9 – 0.14 * 37 kg m/s = 7.9 – 5.18 = 2.72 kg
m/s , so ?, the speed of the bat after hitting the line drive ball = 2.72/0.75 = 3.6 m/s
5. In the figure, a flat piece of metal in the shape of an equilateral triangle, side length = L,
has a circular hole cut out of it in the lower left corner, where the hole just touches (is
tangential to) the two sides as shown. The radius of the hole is L/6. What are the x and y
coordinates (3 sig figs) of the center of mass of the object ? (x=0 and y=0 in the lower left
corner of the triangle as shown.)
Soln.: xcm = (1/Mtotal) * (mtriangle*xcm triangle + mhole*xcm hole) where the mass of the hole is
negative. Since the object is flat, it’s mass is proportional to its area
xcm triangle=L/2
the distance along the x-axis out to the center of the hole is the base of a triangle as shown
below. The angle between the hypotenuse and the base of the little triangle is 30 o, so the
base is found from tan30 =1/sqrt(3) = (L/6) / base ; so the base is L*sqrt(3)/6 = 0.289 L so
xcm hole = 0.289 L; area of circle is pi*(L2/36)=0.08727 L2,
area of triangle = ½ * L * height = ½ * L * sqrt(3) * L/2 = 0.4330 L2
xcm = [1/(.4330 – 0.08727)L2]*(0.4330L2 * L/2 - .08727L2 * 0.289 L)
in the numerator and denominator cancel
; the L2 in the masses
=(1/0.3457)(0.2165 - .02522)L = +0.553 L i. e. to the right of the cm of the triangle without
the hole in its left portion, as expected.
ycm works the same as for the xcm, with ycm hole=L/6 and ycm triangle at the intersection of the
bisectors of all the angles, so tan30 = ycm/(L/2) = 1/(sqrt3) so ycm = L/(2sqrt(3)) = or
sqrt(3)*L/6 or 0.2887 L
so ycm=(1/Mtotal) * (mtriangle*ycm triangle + mhole*ycm hole)
; the L2 in the masses cancels
ycm = (1/0.3457)*(0.433 * 0.2887 L – 0.08727 * L/6) = 0.319 L so the cm in the y-direction
moves up due to the hole from its starting 0.2887 L position.