# A block of mass m  is moving on a horizontal frictionless surface with a speed v_0 as it approaches a block of mass 2m which is at rest and has an ideal spring attached to one side.When the two blocks collide, the spring is completely compressed and the two blocks momentarily move at the same speed, and then separate again, each continuing to move.

1. (a) Briefly explain why the two blocks have the same speed when the spring is completely compressed. (3 points)
2. (b) Determine the speed vf of the two blocks while the spring is completely compressed. (3 points)
3. (c) Determine the kinetic energy of the two blocks as they move together with the same speed. (3 points)
4. (d) When the spring expands, the blocks are again separated, and the spring returns its compressed potential energy to kinetic energy in the two blocks. On a graph, sketch a graph of kinetic energy vs. time from the time block m approaches block 2m until the two blocks are separated after the collision. *You can upload an image of this or describe the graph. (3 points)
5. (e) Write the equations that could be used to solve for the speed of each block after they have separated. It is not necessary to solve these equations for the two speeds. (3 points)
1. When the spring is compressed, both blocks have 0 relative velocity, meaning they move at the same speed. This is similar where two objects move at the same speed after an inelastic collision.
2. v_0/3
3. \frac{1}{6} mv_0^2
4. The graph of KE should be a horizontal line followed by a downwards curved dip, followed by a horizontal line at the original position on the KE axis.
5. mv_0 = mv_{f1} + mv_{f2} \quad \text{and} \quad \frac{1}{2}mv_0^2 = \frac{1}{2}mv_{f1}^2 + \frac{1}{2}mv_{f2}^2
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1. When the spring is compressed, both blocks have 0 relative velocity, meaning they move at the same speed. This is similar where two objects move at the same speed after an inelastic collision.
2. v_0/3
3. \frac{1}{6} mv_0^2
4. The graph of KE should be a horizontal line followed by a downwards curved dip, followed by a horizontal line at the original position on the KE axis.
5. mv_0 = mv_{f1} + mv_{f2} \quad \text{and} \quad \frac{1}{2}mv_0^2 = \frac{1}{2}mv_{f1}^2 + \frac{1}{2}mv_{f2}^2

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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g}
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters.

1. Start with the given measurement: \text{5 km}

2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}

3. Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}

4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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