1. In the diagram below, two blocks of mass M are connected by a string hung over a pulley of mass m. The pulley is a solid disk. The left block is connected to a spring of spring constant K. The surface that the blocks are on has a coefficient of kinetic friction . Determine the speed of the blocks when the upper block moves a distance L to the right. Your answer should be expressed in terms of the given quantities and g.
   
   

2. In the diagram below, two masses M₁ and M₂ are connected by a string hung over a fixed pulley. The pulley has radius R. There are three symmetrically placed holes in the pulley each of radius r and the centre of each is located 2r from the centre of the pulley. Before the holes were drilled in the pulley, it had a mass M. Your answers should only depend on the given quantities and g.
   (a) Obtain an expression for the moment of inertia of the pulley about its axis of rotation.
   (b) Determine an expression for the acceleration of the blocks (Use I for the inertia of the pulley if you do not find an answer for part a).
   
   

3. In the diagram below, a block A of mass 3.00 kg is moving to the right when it collides with block B which has mass 4.50 kg and is moving at 1.00 m/s to the left. The tabletop is frictionless and 1.20 m above the floor.
   (a) If the collision is perfectly elastic, block B lands 1.78 m from the edge of the table. Determine the velocities of each block just before and just after the collision.
   (b) If the collision is totally inelastic, and block A has the initial velocity determined in part (a), find where the blocks land.
   
   

4. In the diagram below, a toy car has an unknown speed v at point A. It is known that the car just makes it (i.e. almost loses contact) around the inside of a loop-the-loop (Point B) of radius R = 0.40 m. It then rolls up an incline to a height h (Point C) where it comes to a complete rest. Determine the initial speed v and the height h. All work must be shown. The surfaces are frictionless.
   
   

5. Two small pieces of silly putty slide across a frictionless surface and collide with, and stick to, a long thin rod. The rod has a length L = 2.20 m and a mass of 0.150 kg. The rod has a fixed axis of rotation at its centre but before the collision it is motionless. The first piece of silly putty has a mass of 0.050 kg and a speed of 6.75 m/s. It will collide with the top edge of the rod. The second piece of silly putty has a mass of 0.035 kg and a speed of 5.00 m/s. It will hit the rod exactly between its centre and lower end. Determine the angular velocity of the system after the collision. Hint - be careful with signs and treat putty as point masses!
   
   

6. An block of mass 0.550 kg is attached to a horizontal spring. The motion of block as a function of time (metres versus seconds) is given by the graph below.
   (a) Write out the expression for x(t). Hint - be very careful choosing φ₀.
   (b) Write out the expression for v(t).
   (c) Write out the expression for a(t).
   (d) What is the spring constant?
   (e) What is the total energy of the system?
   (f) Define resonance.
   
   

Formulas

Error Propagation

Adding or Subtracting	Δ(A+B-C)= ΔA + ΔB + ΔC
Multiplying or Dividing	Δ(AB/C)= (AB/C)(ΔA/A + ΔB/B + ΔC/C)
Powers and Roots	Δ(Az) = zAz-1ΔA
Special Fuctions	sin(θ±Δθ) = sin(θ)±Δθ cos(θ) cos(θ±Δθ) = cos(θ)±Δθ sin(θ) tan(θ±Δθ) = tan(θ)±Δθ /cos2(θ) e(x±Δx) = ex±Δxex ln(x±Δx) = ln(x)±Δx/x
	Note: Δθ must be stated in radians!

Kinematics

vaverage = Δx/Δt	ωaverage = Δθ/Δt
vaverage = (vf+v0)/2	ωaverage = (ωf+ ω0)/2
aaverage = Δv/Δt	αaverage = Δω/Δt
Δx = vaveraget	Δθ = ωaveraget
Δx = v0t + ½at2	Δθ = ω0t + ½αt2
v = v0 + at	ω = ω0 + αt
v2 = (v0)2 + 2aΔx	ω2 = (ω0)2 + 2αΔθ

Translation <-> Rotation

s = Rθ

vtan = Rω

atan = Rα

ac = v2/R

Newton's Laws

ΣFx = max	ΣFy = may	fmax static = μsN	fkinetic = μkN
F = Gm1m1/R2	g(R) = GMplanet/R2	v = [GMplanet/R]½	T2 = 4π2R2/ GMcentral
G = 6.672 × 10-11 N-m2/kg2		g = 9.81 m/s2

Rolling

v_linear = v_tangential

Torque

τ = Iα

τ = r × F

τz = xFy - yFx

Moment of Inertia

Itotal = I1 + I2 + I3 ...

I = Icm + Md2

Work and Energy

	Klinear = ½mv2	Krotational = ½Iω2
Ugravity = mgh	Uspring = ½Kx2

Collisions

p = mv		Pf = Pi
v1f - v1f = -(v1i - v2i)	I = Δp = FaverageΔt

Angular Momentum

L = r × p	Lz = xpy - ypx = rpsinφ	Llinear = bmv
Lorbital = r2mω	LA = IAωA	Ltotal = L1 + L2 + L3 ...
Lfinal = Linitial

Simple Harmonic Motion

Quadratic Formula

if ax²+bx+c = 0, then

Questions? mike.coombes@kwantlen.ca