Notes

5.1 – Momentum & Impulse

Momentum • Momentum is a vector describing how difficult it is to stop a moving object • Total momentum is the sum of individual momenta • p ⃗=mv ⃗ • Units are kg·m/s or N·s Example 1: Changing Momentum • An Aichi D3A bomber mass (3600 kg) departs from its aircraft carrier with a velocity of 85 m/s due east. What is its momentum? • After it drops its payload, its new mass is 3000 kg and it attains a cruising speed of 120 m/s . What is its new momentum? Impulse • As you observed in the previous problem, momentum can change • A change in momentum is known as an impulse (J) • J ⃗=Δp ⃗ Example 2: Impulse • The D3A bomber, which had a momentum of 3.6e5 kg·m/s, comes to halt on the ground. What impulse is applied? Relationship Between Force and Impulse • F ⃗=ma ⃗=m (dv ⃗)/dt=d/dt (mv ⃗ )=(dp ⃗)/dt Example 3: Force from Momentum • The momentum of an object as a function of time is given by p=kt2, where k is a constant. What is the equation for the force causing this motion? Impulse-Momentum Theorem Example 4: Impulse-Momentum • A 6-kg block, sliding to the east across a horizontal, frictionless surface with a momentum of 30 kg·m/s, strikes an obstacle. The obstacle exerts an impulse of 10 N·s to the west on the block. Find the speed of the block after the collision. Example 5: Water Gun • A girl with a water gun shoots a stream of water than ejects 0.2 kg of water per second horizontally at a speed of 10 m/s. What horizontal force must the girl apply on the gun in order to hold it in position? Impulse from F-t Graphs • Impulse is the area under a Force-time graph • Impulse is equivalent to a change in momentum Example 6: Impulse from Force • A force F(t)=t3 is applied to a 10 kg mass. What is the total impulse applied to the object between 1 and 3 seconds?

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5.2 – Conservation of Linear Momentum

Collisions and Explosions • In the case of a collision or explosion, if you add up the individual momentum vectors of all the objects before the event, you'll find that they are equal to the sum of momentum vectors of the objects after the event • Written mathematically, the law of conservation of linear momentum states • p ⃗_initial=p ⃗_final Solving Momentum Problems 1. Identify all the objects in the system 2. Determine the momenta of the objects before the event. Use variables for any unknowns 3. Determine the momenta of the objects after the event. Use variables for any unknowns 4. Add up all the momenta from before the event and set equal to the momenta after the event 5. Solve for any unknowns Types of Collisions • Elastic collision ○ Kinetic energy is conserved • Inelastic collision ○ Kinetic energy is not conserved Example 1: Traffic Collision • A 2000-kg car traveling 20 m/s collides with a 1000-kg car at rest. If the 2000-kg car has a velocity of 6.67 m/s after the collision, find the velocity of the 1000-kg car after the collision Example 2: Collision of Two Moving Objects • On a snow-covered road, a car with a mass of 1100 kg collides head-on with a van having a mass of 2500 kg traveling at 8 m/s • As a result of the collision, the vehicles lock together and immediately come to rest. • Calculate the speed of the car immediately before the collision Example 3: Recoil Velocity • A 4-kg rifle fires a 20-gram bullet with a velocity of 300 m/s. Find the recoil velocity of the rifle Example 4: Atomic Collision • A proton (mass=m) and a lithium nucleus (mass=7m) undergo an elastic collision as shown below. • Find the velocity of the lithium nucleus following the collision Example 5: Collisions in Multiple Dimensions • Bert strikes a cur ball of mass 0.17 kg , giving it a velocity of 3 m/s in the x-direction. When the cue ball strikes the eight ball (mass=0.16kg), previously at rest, the eight ball is deflected 45 degrees from the cur ball's previous path, and the cue ball is deflected 40 degrees in the opposite direction. Find the velocity of the cue ball and the eight ball after the collision 2001 Free Response Question 1 2002 Free Response Question 1 2014 Free Response Question 1

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5.3 – Center of Mass

Center of Mass • Real objects are more complex than theoretical particles • Treat entire object as if its entire mass were contained at a single point known as the object's center of mass (CM) • Center of mass is the weighted average of the location of mass in an object Find CM by Inspection • For uniform density objects, CM is the geometric center • For objects with multiple parts, find CM of each part and treat as a point • For irregular objects, suspend object from two or more points and drop a plumb line. The line intersects at the center of mass Calculating CM for Systems of Particles • r_CM=(∑▒〖mr ⃗ 〗)/(∑▒m) • x_CM=(m_1 x_1+m_2 x_2+…)/(m_1+m_2+…) • y_CM=(m_1 y_1+m_2 y_2+…)/(m_1+m_2+…) Example 1: Center of Mass (1D) Example 2: CM of Continuous System Example 3: Center of Mass (2D) Finding CM by Integration • For more complex objects, you can find the center of mass by summing up all the little pieces of position vectors multiplied by the differential of mass and dividing by the total mass • r ⃗_CM=(∫▒〖r ⃗dm〗)/M Example 4: CM of a Uniform Rod Example 5: CM of a Non-Uniform Rod • Find the center of mass of a non-uniform rod of length L and mass M whose density is given by λ=kx Center of Mass Relationships Center of Gravity • Center of Gravity refers to the location at which the force of gravity acts upon an object as if it were a point particle with all its mass focused at that point • In a uniform gravitational field, Center of Gravity and Center of Mass are the same • In a non-uniform gravitational field, they may be different 2004 Free Response Question 1

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6.1 – Uniform Circular Motion

Uniform Circular Motion • Object travels in a circular path at constant speed • Distance around the circle is its circumference ○ C=2πr=πd • Average speed formula from kinematics still applies ○ v ̅=d/t=2πr Frequency • Frequency is the number of revolutions or cycles which occur each second • Symbol is f • Units are 1/s, or Hertz (Hz) • f=(number of cycles)/second=(number of revolution)/second Period • Period is the time it takes for one complete revolution, or cycle. • Symbol is T • Unites are seconds (s) • T = time for 1 cycle = time for 1 revolution Frequency and Period • f=1/T • T=1/f Centripetal Acceleration • Is an object undergoing UCM accelerating? • Magnitude of Centripetal Acceleration ○ a_c=v^2/r Centripetal Force • If an object is traveling in a circle it is accelerating toward the center of the circle • For an object to accelerate, there must be a net force • We call this force a centripetal force (F_c) • F_c=(mv^2)/r

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7.1 – Rotational Kinematics

Radians and Degrees • In degrees, once around a circle is 360° • In radians, once around a circle is 2π • A radian measures a distance around an arc equal to the length of the arc's radius • Δs=C=2πr Linear vs. Angular Displacement • Linear position / displacement given by Δr or Δs • Angular position / displacement given by Δθ • s=rθ • Δs=rΔθ Linear vs. Angular Velocity • Linear speed / velocity given by v ⃗ • Angular speed / velocity given by ω ⃗ • v ⃗=(ds ⃗)/dt • ω ⃗=(dθ ⃗)/dt Direction of Angular Velocity Converting Linear to Angular Velocity Linear vs. Angular Acceleration • Linear acceleration is given by a ⃗ • Angular acceleration is given by α ⃗ • a ⃗=(dv ⃗)/dt • α ⃗=(dω ⃗)/dt Kinematic Variable Parallels Variable Translational Angular Displacement Δs Δθ Velocity v ⍵ Acceleration a ⍺ Time t t Variable Translations Variable Translational Angular Displacement s=rθ θ=s/r Velocity v=rω ω=v/r Acceleration a=rα α=a/r Time t=t t=t Kinematic Equation Parallels Centripetal Acceleration Example: Wheel in Motion • A wheel of radius r and mass M undergoes a constant angular acceleration of magnitude ⍺. • What is the speed of the wheel after it has completed on complete turn, assuming it started from rest? 2003 Free Response Question 3 2014 Free Response Question 2

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