2010 F=ma Solutions. that is

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1 2010 F=ma Solutions 1. The slope of a position vs time graph gives the velocity of the object So you can see that the position from B to D gives the steepest slope, so the speed is the greatest in that area. 2. Speed is the magnitude of velocity and as we are given a velocity v. time graph, we are looking for the maximum and minimum points and on this graph we see they occur at points B and D and as the magnitude of the velocity are equivalent at those two points, the greatest speed is at points B and D. 3. The area under an acceleration vs time gives the velocity, so from the choices listed we see that the area under the curve at B gives the greatest area. After time B, the speed decreases because the object reverses direction so the overall area will decrease. 4. The time for the piano to hit the ground is. The movers notice the piano after it falls so the piano has to fall another and the time for that is so 5. Define the initial launch point at so we are looking at the time it takes for the projectile to cross this point again as that is the difference in time or 6. At the point of maximum height, because the object reverses vertical direction, so using in the vertical direction, The total time of motion is therefore twice this or Horizontal velocity is constant so. For maximum height, the projectile is in free fall, so ( ) Therefore 7. Angular momentum is defined as so we need to solve for Centripetal force withstands the force from the carousel of so so 8. The only forces acting in the x direction are friction and the horizontal component of gravity. Friction force prevents the car from accelerating up the hill, so it is greater than the horizontal component of gravity. Therefore, from the free body diagram, ( ) 9. The only acceleration is in the x-direction and the forces acting on the ball are at an angle and gravity down, so writing the vertical and horizontal net forces,

2 10. Because block 1 slips on block 2, there is friction on block 1 acting right (block 1 slips left) so by Newton s third law, the same friction acts on block 2 left. We also have the friction from block 2 being pulled right, so summing the forces, 11. Tension is uniform through this one string, so from the first point mass, the forces are T up and gravity down, so. Now let the horizontal distance from the center mass to the left string be Therefore, the horizontal distance from the center mass to the right string is. The net torque of this system is zero, so we can find by taking the torque about the center mass, so Now we find the angle formed from the string about the center mass and the horizontal by taking the net vertical force of the center mass, so and from earlier, so. Therefore, we have a right triangle on both sides of the center mass with legs and, so 12. Apply the work kinetic energy theorem. The work due to gravity is the only work over this time so Therefore for the block to gain the ball must lose in potential energy. The initial energy of the ball is and so 13. The initial PE is equal to the total kinetic energy including rotational energy, so ( ) When the ball is projected up, it has a velocity of and rises to with no rotational energy so, maximum PE = maximum KE so 14. We find the velocity of the 5kg block, call that block a, by using the work kinetic energy theorem, from the onset to right before block a hits the second block, b. The work is due to friction, so. This is the speed right before collision so call it Now this is an elastic collision, so momentum and kinetic energy are conserved, so call the velocity of block b right after the collision to be and the velocity of block a after the collision to be So we have Solve the first equation for and and plug it into the energy equation to get Now apply the work kinetic energy theorem again except and so ( )

3 15. There are no external forces, so momentum and kinetic energy are conserved. First thing is to solve for the speed of the combined masses by using conservation of momentum, so. Energy is conserved and we have the initial kinetic energy of and the kinetic energy of the combined masses and the potential energy of the small mass on the larger mass, or 16. This is an elastic collision, so momentum and kinetic energy are conserved before the collision and while the blocks separate again, so we have. The negative sign indicates the smaller block is moving left and is the mass of the larger block moving right. Energy is conserved, so and from the first equation, and substitute that in the second equation to get ( ) ( ). Now complete the square or use the quadratic equation to get and ( ) from which the two solutions are however we already defined as positive by the inclusion of the negative sign because it s moving left, so indicates the block is moving right, therefore the answer is 17. Each mass is connected to the other masses, so the combined gravitational energy is the sum of the individual energies without overlap, so since there are 4 pairs, there are 6 possible pairs, so the total energy is 18. Force is defined as the derivative of the potential energy or so the corresponding force graph is found by calculating the individual slopes of the line segments on the potential energy graph, and doing so, in order, gives and 0 constant over the appropriate distances, so the correct graph is 19. For this question, compare the positioning in each graph with the energy diagram to determine which ones are possible. For graph 1, x is constant at 15 m so v=0 and on the energy diagram, at x=15 m, so that means the kinetic energy is either at a maximum or 0 so, graph 1 is possible. For graph 2, x is constant at -5m, so v=0 and K=0, but on the energy diagram, U is decreasing so by conservation of energy, there is KE increasing so graph 2 is not possible. For graph 3, x increases linearly from 10 to 15m, so

4 there is constant velocity. From graph 1 we deduced that since U is 0 at m, KE is at a maximum or 0, so graph 3 is reasonable because the KE must be at a maximum. Therefore I and III are possible. 20. Recall conservation of energy or. For this question, we need to find where the potential energy is at a maximum so that K is 0. The kinetic energy is zero when the velocity is zero, so the position when the velocity is zero can be deduced by the maximum and minimum of the position graph, since velocity is the derivative or slope of the position graph, which is zero at a maximum and minimum. Therefore, the maximum and minimum of the position graph is 5 and -5 m, and the corresponding potential energy is -5 J, which is the same for both positions as expected. 21. This problem implies that the gravitational self potential energy depends on the universal gravitational constant, the radius of the object, the density, and some numerical constant. To solve this problem, we need to figure out how energy scales with the radius by determining the dimension quantities of this equation. Energy is in terms of, density is in terms of radius is in terms of and the gravitational constant is in terms of Now let Therefore our equation is where k is the numerical constant. Each quantity increases exponentially by some constants, a, b and c. Therefore, by matching the exponents, our 3 equations are. Therefore, 2 and, so the radius increases exponentially as so increasing the radius by a factor of while keeping everything else constant increases the energy by a factor of so the new energy is 22. Helium is less dense than air (air is mostly made up of nitrogen, which is heavier than helium), so as the car is traveling around the circle, air will move outwards, towards B, so as helium is less dense, the balloon will be pushed inwards, towards D. 23. Because the net force on each tube is the same, the applied forces to each tube are equal, so we know that For the left tube, the change in momentum is equal where is the mass of the flowing water or where is the mass flow rate (mass of water moving per unit time) so so equating the forces gives

5 24. The moment of inertia of a uniform solid disk is A circular section with radius is removed, so the cross sectional area of the disk is removed, so the mass of the removed section is The axis of rotation is units from the center of the removed section, so by the parallel axis theorem, the inertia of the removed section is. Therefore, the moment of inertia of the remaining disk is 25. First find the speed of the ship in the circular orbit of radius by equating the centripetal and gravitational force or. Now in elliptical orbit, energy is conserved so at the minimum and maximum radius, the respective velocities are and. By conservation of energy, we have Angular momentum is conserved in orbit so Plugging in this speed gives Finally,