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Physics Chapter 7
| Question | Answer |
|---|---|
| Linear Momentum (p) | Defined as the product of an object's mass (m) and its velocity (v). Momentum is a vector quantity and the direction of the momentum vector is the same as the velocity vector. Average p = m * average v. |
| What's the unit of linear momentum? | kg * m/s |
| Favr = | change in P over change in T, useful for the conservation of momentum |
| Impulse | Defined as the product of force (F) acting on an object and the time (delta T) during which the force acts. I = F * delta t. An impulses causes a change in an object's momentum. |
| What is the unit for impulse? | N * s, where 1 N * s = 1 kg * m/s |
| Graphically, how is impulse represented? | It can be represented by the area under the time and force graph. |
| What does delta T refer to? | Delta t refers to the time it takes for an object to complete its action (collision), which is usually very short. |
| Conservation of Linear Momentum | means that initial momentum has to equal final momentum (Pi = Pf). Also means Pf -Pi = 0. |
| Main Condition for the Conservation of Linear Momentum | If you want delta P = 0, then Fnet must equal 0. |
| Remember that during collisions, the ENTIRE system will have its momentum conserved but usually not individual parts. Here's how you would write the equation for that: | m1v1 + m2v2 = m1v1f + m2v2f |
| Elastic Collisions | Objects do not stick together, both the momentum and the total energy of the system are conserved. So: Pi = Pf ETi = ETf KEi + PEi = KEf + PEf -> KEi = KEf |
| Inelastic Collisions | Collisions that can lead to damage so the total energy can no longer be conserved. Only momentum is conserved. Pi = Pf. |
| Completely Inelastic Collisions | Objects collide and then stick together, Momentum is conserved but kinetic energy is not, because kinetic energy is converted into heat or sound. m1v1 + m2v2 = (m1 + m2)vf |
| Complete Lecture Problem 1 | -- |
| Complete Lecture Problem 2 | -- |
| Head-On Collision | One dimensional collision in which the objects' velocities are all lined up along the same line, which remains true before and after the collision. Both momentum and kinetic energy are conserved. |
| When the two masses are equal... | m1 will come to a perfect stop after the collision. |
| When m1 is larger than m2 | m1 moves forward after collision |
| When m1 is smaller than m2 | m2 bounces back after collision. |
| Complete Lecture Problem 3 | -- |
| Head-On Collision Problem Equations (when they remain along the same line) for V1f | ((m1-m2)v1) / (m1+m2) |
| Head-On Collision Problem Equations (when they remain along the same line) for V2f | (2m1v1) / (m1 + m2) |
| Complete Lecture Problem 4 | -- |
| Ratio of Masses in Inelastic Collisions | If an object is initially at rest and explodes into two pieces, expect the smaller piece to have 2 times the speed of the larger one if the larger is 2m compared to m. |
| Complete Lecture Problem 5 | -- |
| At the highest point, there is... | Zero velocity |
| When dealing with the pendulum and bullet scenario, it is best to... | Begin with a conservation of kinetic energy equation and carry on into conservation of momentum. |
| Center of Mass | Point where the mass of an object can be considered to be concentrated |
| Center of Mass & Size | If the object is small, there is no difference between the center of mass and the center of gravity. If the object is mounted horizontally, the center of mass and the center of gravity are identical. |
| Center of Mass & Vertical Mounting | If you mount it vertically, the center of mass will continue to be the center of the object but the pull of gravity will difference with distance. Gravity decreases with altitude - the pull on the lower part, so you cannot completely balance the rod. |
| Main Difference Between CM and CG in Vertical Mounting | The center of gravity has to shift down and is dependent on gravity while center of mass really only relies on the mass itself. |
| Complete Lecture Problem 6 | -- |
| Complete Recitation Problem 1 | -- |
| Complete Recitation Problem 2 | -- |
| Complete Recitation Problem 3 | -- |
| Homework Problem 1 | -- |
| Homework Problem 2 | -- |
| Homework Problem 3 | -- |
| Homework Problem 4 | -- |
| Homework Problem 5 | -- |
| Homework Problem 6 | -- |
| Homework Problem 7 | -- |
| Homework Problem 8 | -- |
| Homework Problem 9 | -- |
| Homework Problem 10 | -- |
| Remember that velocity can be found from height with the following equation | Square root of 2gh |
| MEGA IMPORTANT TIP | Always remember to put the mass in when you're doing momentum calculations! No silly mistakes! |
Created by:
smurtab