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Family Name: Given Name: Student number: Academic Honesty: In accordance with the Academic Honesty Policy (T0.02), academic dishonesty in any form will not be tolerated. Prohibited acts include, but are not limited to, the following: making use of any books, papers, electronic devices or memoranda, other than those authorized by the examiners. speaking or communicating with other students who are writing examinations. copying from the work of other candidates or purposely exposing written papers to the view of other candidates. Question # Mark Maximum Mark Multiple Choice 20 0 2 0 3 0 4 0 Total = /60 Enter your answers to the multiple choice questions here by blackening in the circle corresponding to the best answer. There is only one correct answer per question.. A B C D E 2. A B C D E 3. A B C D E 4. A B C D E 5. A B C D E 6. A B C D E 7. A B C D E 8. A B C D E 9. A B C D E 0. A B C D E

There are 0 multiple choice questions. Select the correct answer for each one and mark it on the bubble form on the cover sheet. Each question has only one correct answer. (2 marks each). Three boxes are connected by massless strings and are on a frictionless table accelerating to the right. Each box has the same mass of 5 kg. How does the tension in the three strings compare? T 3 (a) T < T 2 < T 3 (b) T = T 2 = T 3 T 2 (c) T > T 2 > T 3 Correct (d) T = T 2 < T 3 (e) T = T 2 > T 3 All the boxes are accelerating together to the right. The T 2 string is only pulling the last two boxes and the T string is only pulling one box. According to Newton s second law, if the accelerations are the same then the tensions are less for the lower masses. 2. A box is sitting on the floor of a train car that is travelling at constant speed. What best describes the net force on the box? (a) The net force is mainly static friction. (b) The net force is mainly kinetic friction. (c) The net force is both kinetic friction and static friction. (d) The net force is the normal force. (e) The net force is zero. Correct If the velocity is constant velocity then the net force must be zero. 3. An object of mass m is moving in uniform circular motion with radius R and speed v. If the centripetal force on the mass is doubled and the radius of rotation is doubled, how does the speed change if it is to maintain uniform circular motion? (a) v stays the same (b) v decreases. (c) v quadruples (d) v increases by about 40% (e) v doubles T F = mv 2 /r, therefore v = Fr/m, and v 2 = (2F)(2r)/m = 2 Fr/m = 2v 4. A box of mass M sits on a horizontal table. A string with tension T pulls to the right, but static friction between the box and the table prevents the box from moving. What is the magnitude of the static frictional force acting on the box? (a) Mg (b) µ s Mg (c) T correct While the box is stationary, the static frictional force must equal the tension on the string. µ s Mg gives the maximum value that the static friction force can be but it can be less than that. 5. A box is initially sliding across a frictionless floor toward a spring which is attached to a wall. The box hits the end of the spring and compresses it, eventually coming to rest for an instant before bouncing back the way it came. The work done by the spring on the box as the spring compresses is: (a) Zero (b) Positive (c) Negative correct During compression the displacement is opposite the direction of the force so work is negative. 6. A circus performer of weight W is walking along a high wire as shown. The tension in the wire (a) depends on whether he stands on one foot or two feet (b) is approximatelyw 2

(c) is approximately W/2 (d) is much less that W (e) is much more than W correct The weight of the performer is in equilibrium with the vertical components of the tension of the rope on the right side and left side: mg = 2T y. When the angle of the rope is small, the magnitude of the tension must be large in order for the vertical component to be sufficiently large to balance the weight: T = mg/(2 sin θ). 7. In Case, a mass M hangs from a vertical spring having spring constant k and is at rest in its equilibrium position. In Case 2 the mass has been lifted a distance D vertically upward. If we define the potential energy in Case to be zero, what is the potential energy of Case 2? (a) 2 kd2 (b) MgD (c) 2 kd2 + MgD At equilibrium it requires work to either push the mass either up or down so the minimum of the potential must be at the equilibrium position. If y is the position relative to the new equilibrium then y = y D was the position of the end of the spring before the mass was hung on it. Expressing the energy in the original coordinate system, note that 2 ky2 = 2 ky 2 kdy + 2 kd2. kdy = mgy thus the gravitational potential is included in the new coordinate system. 8. Here is a graph of potential energy as a function of position of an object moving on the x axis. At which point is the force greatest in the negative direction? U(x) The force is the negative of the first derivative of the potential with respect to the spatial coordinate. F(x) = du dx so point B is the answer. 9. In case A a block is pushed by force F for second. In case B a block of twice the mass is pushed by the same force F for 2 seconds. In both cases the track is frictionless and the blocks are initially at rest. Compare the magnitudes of the final momenta of the two blocks: (a) p A < p B (b) p A > p B (c) p A = p B The force times time, is the impulse, and equals the change in momentum. Thus the momentum change is the block B. Block B gains twice as much momentum as block A. 0. Three plane boards of equal thickness and mass have been cut into equilateral triangles. Which of the two arrangements has the higher centre of mass? (a) a (b) b (c) same a b 0 a b c d e x The c.m. of each triangle is at the centroid, /3 of the way up from the base to the tip. Case a has a higher c.m. because 4/3+4/3+/3 > 2/3+2/3+4/3. 3

There are three written problems. Show all your work and explain your reasoning to get full credit.. The KC-35 Stratotanker, also known as the Vomit Comet, is used by NASA to produce a weightless (or microgravity ) environment for brief periods of time. It takes the aircraft 25 s to travel from point A to point B along its flight trajectory. Assume the speed of the aircraft is constant. (a) On the left-hand side of the picture, draw a free body diagram of the passenger in the airplane when it is before point A while the airplane is still climbing in a straight line with constant speed at 45 to the horizontal. The passenger is 00 kg, sitting on the floor in the airplane. g F normal F friction A B R F gravity 45 45 F gravity (b) Draw a free body diagram of the passenger in the airplane when it is at the top of its trajectory. (right-hand box) Assume in this case that the passenger is weightless and floating in the middle of the cabin. (c) If the normal force on the passenger is zero, what is the radius, R, of the trajectory assuming uniform circular motion? At the top of the trajectory the centripetal acceleration must be equal to g in order for there to be no non-gravitational force on the passenger. v 2 R = g Also the speed is given by the distance travelled during /4 circle divided by the time from A to B. Combining and solving for R v = πr 2T R = 4T 2 g π 2 = 2482 m (If one approximates π 2 g 0 the answer is 2500 m. (d) If the airplane takes the same path but at a slower speed so that it takes 30 s to go from A to B, what is the normal force on the passenger at the same radius R? For ver B F normal = 74 N v = πr 2T = 59.8 m/s a c = v2 = 6.80 m/s2 R F normal = m(a c g) = 300 N 4

(e) Explain what would happen to the passenger if the plane went faster and took 20 s to go from A to B at the same radius R. The passenger would seem to float up and press against the ceiling in the cabin as the plane accelerates downward faster than g. 5

2. A block having mass m slides on a frictionless surface and runs into a block of mass m 2 which is initially at rest. The blocks stick together and the ratio of the initial speed of the block to the speed of blocks after the collision is r = v i /v f. In the following, answer questions with a formula using r, m and m 2. (a) Find the formula for m 2 in terms of m and r. m v = (m + m 2 )v 2 For ver B: m 2 = m (r ) m = m 2 /(r ) (b) Verify that your expression in (a) is correct for the cases m = m 2 and m >> m 2 Case : r = 2, m 2 = m (2 ). Case 2: r =, m 2 0 m which means that m 2 << m. (c) What is the fraction of the kinetic energies after the collision to that before the collision, K f /K i? K f K i = 2 (m + m 2 )v 2 f 2 m v 2 i (d) What fraction of the initial momentum remains after the collision? The ratio of momentum after to before is because the total momentum remains constant if there are not external forces on the system. = r 6

3. A ball of mass 0.06 kg hits a racquet with velocity 20 m/s. While in contact with the racquet, the force on the ball is opposite the direction of impact velocity and is measured at 2 ms intervals as shown in the table below. (a) Make a graph of F vs t. t (ms) F (N)! 0! 0! 2! 00! 4! 300! 6! 400! 8! 300! 0! 00! 2! 0 (b) What is the approximate impulse on the ball? (Explain how you got it!) The magnitude of the impulse is J = 2 0 3 (50 + 200 + 350) 2 = 2.4 N s (c) Predict the final velocity of the ball. v = p/m = 2.4/0.06 = 40 m/s. v f = 20 m/s. (Reverse the sign of J to account for the direction of the force.) (d) How much work was done on the ball by the racquet? W = K = 2 mv2 i 2 mv2 f = 0 7

4. Two earth-like planets are in space a distance 2a apart. The masses and radii of both are the same: M and R. A spaceship, mass m is sent on a mission from one planet to the other. (a) Position the planets on the x axis with the origin half-way between the planets. The distance of the centre of each planet from the origin is a. Write the equation for the potential energy U(x) of the space ship at any point between the surfaces of the two planets as a function of x and the parameters a, M, m and R. Assume that U = 0 very far away from both planets. Add the potentials of the two planets. Make sure the potentials are negative and make sure the denominators correctly represent the distances between the planets and the spaceship and are positive. ( U(x) = GMm x + a + ) a x (b) Now assume that M = 5 0 24 kg, R = 6000 km, a = 25R and m = 2000 kg. G = 6.67 0 N m 2 /kg 2 Sketch the potential energy as a function of x between the surfaces of both planets. (c) If the spaceship leaves from the surface of the planet whose centre is at a what is the smallest initial velocity needed in order to arrive at the surface of the other planet? The energy needed to get from one planet to the other is just the energy to go from the surface of one planet to the midpoint: U(24R)= -3.435 GJ, U(0)=-8.893 GJ U = 04.5 GJ. v = 2 U/m = 0223 m/s. For ver B v = 0295 m/s (d) What would be the speed of the spaceship as it approaches the other planet a distance R above its surface? (Assume free fall.) 2(U(0) U(23R)) v f = m = 7000.4 m/s. For ver B,v = 76 m/s 8