PHY131 Ch 2 to 4 Exam Name

#1 A lump of wet putty moves in uniform circular motion as it rides at a radius of 20.0 cm on the rim of a wheel rotating (vertical) counterclockwise with a period of 5.00 msec. The lump then happens to fly off the rim at an angle 60° above the horizontal. It leaves the rim at a height of h = 2.00 m from the floor and at a distance d = 3.00 m from a wall. At what height on the wall does the lump hit? /
|v| = dist / t
|v| = 2 π r / T
|v| = 2π(0.2) / 0.005
|v| = 251 m/s 10/10 / vx = v cosθ; vx = Δx / Δt
v cosθ = Δx / Δt
251 cos60° = 3 / Δt
t = 0.0239 sec 10/10 / 5/5 5/5 5/5
y = ½ a t2 + vy t + yo
y = ½-10t2 + v sin60° t + 2
y = 7.19 m
#2 The acceleration of a particle moving on a horizontal plane is (a = 6t). At t = 0, the position vector of x = 6 m is the location of the particle, which then has the velocity vector, v = 7.00 m/s. At t = 4.00 s, what is its position vector. (SI Units) / 6t = dv / dt
6∫t dt = ∫dv
3t2 + 7 = v
8/8 8/8 / 3t2 + 7 = dx/dt
3∫t2dt + 7∫dt = ∫dx
8/8 8/8
t3 + 7t + 6 = x
43 + 7(4) + 6 = x
x = 98 meters 3/3
#3 Same as above problem, but the particle is now moving in the horizontal xy plane with a = 6t i + 12t j m/s2. Also at t = 0, the position vector of r = 6i – 15j meters is the location of the particle, which then has the velocity vector, v = 7.00i + 32j m/s. At t = 4.00 sec, what is the angle between its direction of travel and the positive direction of the x axis? / 12t = dv / dt
12∫t dt = ∫dv
6t2 + 32 = v
6t2 + 32 = dy/dt
6∫t2dt+32∫dt = ∫dy
2t3 + 32t - 15 = y
2*43 + 32(4) -15 = y
y = 241 meters 7/7 / cos ax = r×bx / |r||b| ; bx = ( 1, 0)
r2 = 982 + 2412; |r| = 260 m 3/3
5/5 5/5 5/5 5/5
cos ax = r · bx / |r| |bx |
cos ax = (98,241) · (1,0) / |260| |1|
ax = 67.9°
OR #3 You a defending a fort on top of a hill. You shoot your water cannon at a 30˚angle above the horizontal with a velocity of 60 m/s. It hits your opponent 8 seconds later. (a) What is the impact velocity of the water balloon? (b) What are the coordinates (x, y) of the target? (416,-80) /
vy_0 = 60 sin30°
vy_0 = 30 m/s
3/3 7/7
a = Δvy / Δt
-10 = (vf - 30) / 8
vy_f = -50 m/s / vx = 60 cos30°
vx = 52 m/s
|v| = (522+502)½
|v| = 72.1 m/s 5/5 / (a)
tan θ = -50/52
θ = -43.9° 5/5
(b) y = ½ a t2 + vy t + yo
y = ½-10 82 + (30) 8 + 0
y = -80 m 5/5
vx = Δx / Δt
52 = Δx / 8
Δx = 416 m 5/5

OR #3 The determined coyote is out once more to try to capture the elusive roadrunner. The coyote wears a pair of Acme jet-powered roller skates, which provide a constant horizontal acceleration of 12 m/s2 (this remains constant until he impacts with bottom of canyon). The coyote starts off at rest 100 m from the edge of a cliff at the instant the roadrunner zips past him in the direction of the cliff. At the brink of the cliff, the roadrunner escapes by making a sudden turn, while the coyote continues straight ahead. If the roadrunner moves with a constant speed

(a) determine the minimum speed he must have to reach the cliff before the coyote. / (b) If the cliff is 100 m above the floor of a canyon, determine where the coyote lands in the canyon / (c) Determine the components of the coyote’s impact velocity.
xrr = ½at2 + vrr t + xo
100 = 0 + vrr t + 0
Time to edge of cliff
t = 100 / vrr 5/5 / Time to edge of cliff
t = 100 / vrr
t = 4.08 sec 2/2 / …to bottom of cliff
100 = ½10 t2
tbottom = 4.47 sec 2/2 / 5/5
a = Δvy / Δt
-10 = (vf - 0) / 4.47
vy_f = -44.7 m/s
xcoy = ½ a t2 + vx t + xo
100 = ½12t2 + 0 + 0
100 = ½12(100 / vrr)2
vrr =24.5 m/s 5/5 / xcoy = ½ a t2
xcoy = ½ 12 (4.08 + 4.47)2
2/2 4/4
xcoy = 440 m / 5/5
a = Δvx / Δt
12 = (vf - 0) / (4.08 + 4.47)
vx_f = 102 m/s