Exam 6 review

For each class, I will select six questions from the Webassign review set and modify them slightly, as in changing the diagram or graph slightly or changing the numbers.  Each question will have four steps:

1. Entering the principle most applicable to the situation

2. Entering the system to which the principle is applied

4. Tying all three together with the explanation.

These are the four principles, which will also be provided at the beginning of the test:

Net external force equals mass times acceleration

Net external work equals change in system energy

Net external impulse equals change in system momentum

Net external force equals system mass times center of mass acceleration

Though not always true, here is some guidance about when a principle applies:

 ∑F = ma To determine the acceleration of an object or to determine an unknown force on an object. ∑W = ΔE To determine the speed or height of an object after some change has occurred. ∑J = Δp When objects collide or fly apart (as in an explosion). ∑F = (msys)(acm) When internal forces rearrange a system; these questions often directly refer to the center of mass concept.

And some general guidance about what belongs in a system; again, not necessarily true in every situation:

 ∑F = ma Whatever has mass and is accelerating.  If a block slides down a stationary ramp, the ramp is not accelerating, so it doesn’t belong in the system. ∑W = ΔE It’s often most convenient to put everything in the system so that the net external work is zero. ∑J = Δp Whatever has mass and is moving in the initial and/or final stage of motion. ∑F = (msys)(acm) Whatever has mass and is changing in position.

Before I show some examples, I should mention a few common errors I’ve noticed in Webassign explanations.

1. Stating that an object moves with a constant velocity or that the center of mass of a system moves with a constant velocity “because there are no external forces on the system”.  The idea that there are no forces on the object or system is really only approximately true in deep space.  Everywhere else, it’s only correct to say, “because the net external force is zero”.  For instance, a puck sliding on ideal ice coasts at a constant velocity not “because there are no outside forces” but because “the net external force (consisting of gravity and the normal force) add to zero.”

2. Not being careful about defining systems when using conservation of energy.  For example, if spring energy is in your equations, the spring must be in the system.  If gravitational potential energy is in your equations, the Earth must be in the system.

3. Not being careful about defining systems when using Newton’s second law.  For example, when a block slides along a frictive surface and you are calculating the acceleration of it slowing down, the block is the system.  Just because the floor is involved in the situation, does not mean the floor is in the system.

Below are some example questions.  Each has the original Webassign question, an example of how I would change it slightly, and then an example of what a good test answer would be.

Original question: Modified question:

 The coefficient of friction between a 120kg crate and the floor is 0.60.  How hard must the crate be pushed so that it slides to the right with an acceleration of 3m/s2?   a. 1200N b. 720N c. 360N d. 1560N e. 1080N

 1. Entering the principle most applicable to the situation   Net external force equals mass times acceleration   2. Entering the system to which the principle is applied   The 120kg crate   3. Choosing the correct answer   e   4. Tying all three together with the explanation.   The sum of the force of gravity and the normal force are zero, so the net external force on the crate is equal to the force of the push plus the force of kinetic friction.  Kinetic friction magnitude is equal to the normal force of 1200N times the coefficient of 0.60 and is negative if the crate is sliding rightward and kinetic friction opposes the sliding, so kinetic friction force is -720N.  Net external force equals mass times acceleration then becomes Fpush + (-720N) = (120kg)(3m/s2) and Fpush= 1080N.

 Choosing the correct principle 1 pt Choosing the correct system 1 pt Choosing the correct answer 1 pt Stating that the normal force is 1200N 1 pt Stating that the force of kinetic friction is –720N 1 pt Stating that the push force plus the force of kinetic friction equals the mass of the crate times the acceleration of the crate 2 pts

With only six questions, you should have enough time to write an overly-thorough explanation for each answer.

Original question: Modified question: 1. Entering the principle most applicable to the situation   Net external work equals change in system energy   2. Entering the system to which the principle is applied   The block   3. Choosing the correct answer   d   4. Tying all three together with the explanation.   The only force inputting work to the system of the block is the force of the spring scale.  This work input then appears as the kinetic energy of the block, ½mv2.  Because work is the area under a force versus position (or distance) graph, the option with the greatest area (d) has the greatest work input to the block.  This implies the greatest kinetic energy for (d) and thus the greatest speed, v.

Original question: Modified question: As shown above, a disc of mass m is moving horizontally to the right with a speed v on a table with negligible friction when it collides with a second disc of mass 2m.  The second disc is moving horizontally to the right with speed v/2 at the moment of impact.  After colliding, the smaller disk rebounds with a velocity of –v/2.  What is the final velocity of the larger disk?   (A) v/2          (B) 5v/4          (C) v          (D) 2v          (E) 3v/4

 1. Entering the principle most applicable to the situation   Net external impulse equals change in system momentum   2. Entering the system to which the principle is applied   The small disk and the large disk   3. Choosing the correct answer   b   4. Tying all three together with the explanation.   The net external force on the system is zero, so the net external impulse input to the system is zero.  Therefore, there is no change in the system’s momentum.  The initial momentum (p = mv) equals (m)(v) + (2m)(v/2) = 2mv and this must equal the final momentum.  If 2mv = (m)(-v) + (2m)(vf), then vf = 5v/4.

Original question: Modified question: A student stands at the center of a raft floating in a pool with equally spaced marks along the bottom.  The student and the raft have the same mass.  The student walks to the right end of the raft.  Which of the following best shows the final locations of the raft and student relative to the marks at the bottom of the pool?  Assume that there is no drag force between the raft and water. 