Phy2048 Spring 2008
Review Exam
• Q20
• Q19
• Q17
• Q16
• Q13
• Q12
• Q10
• Q8
• Q5
• Q2
Q17: The speed of a 0.41 kg hockey puck, sliding across a level ice surface, decreases at
the rate of 0.61 m/s2 . The coefficient of kinetic friction between the puck and ice is:
FN
fk
Fg
v
Forces on the puck:
1) Fy,net = FN - Fg = 0 --> FN = Fg = mg
2) Fx,net = - fk = max
3) fk = μkFN = μk mg
4) -(μk mg) = max --> μk = - max/mg = - ax/g
5) ax = - 0.61 m/s2
μk = - (-0.61 m/s2)/(9.81 m/s2) = + 0.0622
Q19: Block A, with a mass of 10 kg, rests on a 35° incline. The coefficient of static friction is 0.40.
An attached string is parallel to the incline and passes over a massless, frictionless pulley at the
top. The largest mass mB, attached to the dangling end, for which A remains at rest is:
Forces on B:
1) Fy,net = T - mBg = mayB = 0 --> T = mBg
Spring 08, Exam 1: Q19
N
fs
!
T
T
A
B
Fg,A
Fg,B
Forces on A:
2) Fy,net = FN - Fg,A cosθ = mayA = 0 --> FN = mAg cosθ
3) Fx,net = T - fs,max - Fg,A sinθ = maxA = 0
-->T = fs,max + Fg,A sinθ
4) fs,max = μsFN = μs mAg cosθ
5) T = μs mAg cosθ + mAg sinθ = mAg(μscosθ + sinθ)
Back to Eqtn 1)
T = mBg = mAg(μscosθ + sinθ)
6) mB = mA(μscosθ + sinθ)
= (10 kg)[0.4 cos(35°) + sin(35°)]
= 9.0 kg
Q20: A block is suspended by a rope from the ceiling of a car. When the car rounds a 45-m radius
horizontal curve at 22 m/s (about 50 mph), what angle does the rope make with the vertical?
Spring 08, Exam 1: Q20
1) Fx,net = T sin! = max = mv2/r
2) Fy,net = T cos! - mg = may = 0
3) T = mg/cos!
4) mg sin!/cos! = mg tan! = mv2/r
5) tan! = v2/gr = (22 m/s)2/(9.81 m/s2)(45 m)
= 484 m2/s2 / (441.5 m2/s2)
= 1.096
! = atan(1.096) = 47.6o
!
T
R = 45 m
v = 22 m/s
Fnet
Fg
Q16: Two blocks with masses m and M are pushed along a horizontal frictionless surface by a horizontal
applied force F as shown. The magnitude of the force of either of these blocks on the other is:
Spring 08, Exam 1: Q16
F
a)
M
FB,A
b)
m
FA,B
Forces on block A:
1) Fx,net = F - FB,A = MaA
Forces on block B:
2) Fx,net = FA,B = maB
The blocks move together
3) aA = aB == a
Newton’s 3rd law
4) FA,B = FB,A == Fcontact
1) -> F - Fcontact = Ma
2) -> Fcontact = ma
5) F - ma = Ma -> F = (M+m)a -> a = F/(m+M)
6) Fcontact = ma = m/(m+M) F
Q13: A 1-kg mass on a horizontal air track (so we can ignore friction) is attached to a compressed, massless,
horizontal, ideal spring fixed on its other end to an end of the air track and the spring is released. If the mass
initially has an acceleration of 5.6 m/s2 , the force of the spring on the mass has a magnitude of:
FSpring
Forces on the mass:
1) Fx,net = FSpring = max
2) FSpring = (1 kg)(5.6 m/s2) = 5.6 kg m/s2 = 5.6 N
Q12: A ball is thrown horizontally from the top of a 20-m high hill. It strikes
the ground at an angle of 45° . With what speed was it thrown?
vf,x
θ
vf
vf,y
Initial condition: ax = 0, ay = g, v0y = 0, y0 = 20m
Motion in the x-direction:
1) ax = 0
2) vx = v0x + axt = v0x
3) x = x0 + v0xt + ½axt2 = v0xt
7) y-y0 = ½ayt2,
8) tfinal = sqrt[2(y-y0)/ay]
9) vy,final = aytfinal = ay [2(y-y0)/ay] =
2) -> vx,final = v0x
Motion in the y-direction:
4) ay = -9.81 m/s2
5) vy = v0y + ayt = ayt
6) y = y0 + v0yt + ½ayt2 = y0 + ½ayt2
10) tanθ = vy,final/vx,final = 2(y-y0)ay / v0x
Final condition: y = 0
2(y-y0)ay
11) v0x = 2(y-y0)ay / tanθ
= sqrt[2(-20 m)(-9.81 m/s2)] / tan(45°)
= sqrt[392 m2/s2] / 1
= 19.8 m/s
Q10: A plane traveling north at 200 m/s turns and then
travels south at 200 m/s. The change in its velocity is:
Initial velocity: vi = 200 m/s î
Final velocity: vf = - 200 m/s î
∆v = vf - vi = (- 200 m/s î) - (200 m/s î) = -400 m/s î
vf = 200 m/s
vi = 200 m/s
Q8: Let ⃗S = (1m)ˆi + (2m)ˆj + (2m)ˆk and ⃗ T = (3m)ˆi + (4m)ˆk.
The angle between these two vectors is given by:
1) S· T = |S||T| cosθ = SxTx + SyTy + SzTz = (1· 3) + (2· 0) + (2· 4) = 11
2) |S| = sqrt[12 + 22 + 22] = sqrt[9] = 3
3) |T| = sqrt[32 + 02 + 42] = sqrt[25] = 5
4) cosθ = 11 / (3· 5) = 11/15 = 0.733
θ = cos-1 (11/15) = 42.8°
Q5: The coordinate of a particle in meters is given by x(t) = 16t - 3.0t3,
where the time t is in seconds. The particle is momentarily at rest at t =
1) x(t) = 16 t - 3 t3
2) vx(t) = dx/dt = 16 - 9t2
3) vx(trest) = 0 = 16 - 9t2rest
4) t2rest = 16/9 -> trest = 4/3 s = 1.333s
Q2: During a short interval of time the speed v in m/s of an automobile is given by v = at2 + bt3 ,
where the time t is in seconds. The units of a and b are respectively:
1) v (m/s) = at2 + bt3
2) v (m/s) = a[?] (t2 s2) + b[?] (t3 s3) --> a = [m/s3]
b = [m/s4]