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To solve the problem of finding the final velocities of both blocks after an elastic collision, we need to use both the conservation of momentum and conservation of kinetic energy principles. The masses and initial velocities will be plugged into these equations to determine the final velocities.

Let:
m_1 = 1.5 \, \text{kg} (mass of Block 1)
m_2 = 0.75 \, \text{kg} (mass of Block 2)
u_1 = 3 \, \text{m/s} (initial velocity of Block 1)
u_2 = 0 \, \text{m/s} (initial velocity of Block 2, as it is at rest)
v_1 (final velocity of Block 1)
v_2 (final velocity of Block 2)

Step Derivation/Formula Reasoning
1 m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 Conservation of momentum equation. In elastic collisions, momentum is conserved before and after the collision.
2 1.5 \times 3 + 0.75 \times 0 = 1.5 v_1 + 0.75 v_2
4.5 = 1.5 v_1 + 0.75 v_2
Substitute known values
3 \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 Conservation of kinetic energy equation, also conserved in elastic collisions.
4 \frac{1}{2} \times 1.5 \times 3^2 + \frac{1}{2} \times 0.75 \times 0^2 = \frac{1}{2} \times 1.5 \times v_1^2 + \frac{1}{2} \times 0.75 \times v_2^2
6.75 = 0.75 v_1^2 + 0.375 v_2^2
Substitute known values
5 Solve two equations (from previous steps) simultaneously:

Equation 1:  4.5 = 1.5 v_1 + 0.75 v_2

Equation 2:  6.75 = 0.75 v_1^2 + 0.375 v_2^2

Use algebraic methods (substitution, elimination) to solve for v_1 and v_2 from equations from step 2 and step 4.
6 After solving:
v_1 = 1 \, \text{m/s}
v_2 = 4 \, \text{m/s}
Final solution

This solution strategy provides the final velocities of each block after the collision, assuming perfectly elastic conditions where both momentum and kinetic energy are conserved.

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Block 1 (1.5 kg): 1.0 m/s to the right

Block 2 (0.75 kg): 4.0 m/s to the right

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##### Dark Mode Equation Sheet (Download)
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)}
\omega (Angular Velocity)\text{radians per second (rad/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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