α f k θ y N m mg Figure 1 Solution 1: (a) From Newton s 2 nd law: From (1), (2), and (3) Free-body diagram (b) 0 tan 0 then

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Question [ Work ]: A constant force, F, is applied to a block of mass m on an inclined plane as shown in Figure. The block is moved with a constant velocity by a distance s.

1 Question [ Work ]: A constant force, F, is applied to a block of mass m on an inclined plane as shown in Figure. The block is moved with a constant velocity by a distance s. The coefficient of kinetic friction between the inclined plane and the block is μ k. (a) Draw free-body diagrams and find the force F (5 points). (b) What is the net work done by F (0 points)? (c) Check case (a) if α=0 and comment on the result (5 points). Figure Solution : (a) From Newton s 2 nd law: cos sin () sincos (2) x 0, y 0 and (3) From (), (2), and (3) sin cos cos sin cos sin cos sin cos tan (b) 0tan0 then sin cos F α f k θ y N m mg Free-body diagram x The first term is the work done against the gravitation, the second term is the work done against friction.

Question 2 [ Conservation of Energy ]: A block of mass m is released from an inclined plane as shown in Figure 2. The coefficient of kinetic friction between the inclined plane and the block is μ k.

2 Question 2 [ Conservation of Energy ]: A block of mass m is released from an inclined plane as shown in Figure 2. The coefficient of kinetic friction between the inclined plane and the block is μ k. Assume that the spring is initially uncompressed and there is no friction on the horizontal surface. (a) Find the maximum compression in the spring (0 points). (b) If the block is bounced back from the spring, find the maximum height on the inclined plane that the block can reach (7 points). (c) Check case (a) if μ k =0 and comment on the result (3 points). Figure 2 Solution 2: (a) The total initial energy sin () The speed of the block when it reaches the bottom of the inclined plane: 2 sin cos The maximum compression in the spring sin cos N m mg θ f k Free-body diagram x y (b) When the block reaches at maximum height on the inclined plane, 0 2 cos sin sin cos cot (c) If 0, then sin, the block is returned back its initial height (Total energy is conserved).

Question 3 [ Momentum ]: A man of mass m clings to a rope ladder suspended below a balloon of mass M as shown in Figure 3. The balloon is stationary with respect to the ground.

3 Question 3 [ Momentum ]: A man of mass m clings to a rope ladder suspended below a balloon of mass M as shown in Figure 3. The balloon is stationary with respect to the ground. (a) If the man begins to climb the ladder at a speed v (with respect to the ladder), in what direction and with what speed (with respect to Earth) will the balloon move? (2 points). (b) If the man then stops climbing, what is the speed of the balloon (with respect to Earth) (4 points)? (c) Assuming that M >> m, what is the speed of the balloon in case (a) (4 points)? Figure 3 Solution 3: Method : Center of mass (a) The net force on the balloon-man system is zero so the center of mass of this system is constant: cm cm 0 The balloon is moving downward. (b) If 0 0, the balloon is stationary. (c) If 0, / 0 : expected Method 2: Conservation of momentum (a) The momentum is conserved: 0,,

4 Question 4 [ Collision ]: The two balls, m 2 and m 3, on the right of Figure 4 are slightly separated and initially are at rest; the left ball, m, is incident on m 2 with speed v 0. Assuming head-on elastic collisions and no friction; (a) Find speeds of m and m 2 just after v the first collision (8 points). 0 (b) Assuming that m =m 2 =m and m 3 m, show that there are two collisions and find all final velocities (6 points). (c) Assuming that m =m 2 =m and m 3 > m, Figure 4 show that there are three collisions and find all final velocities (6 points). m m 2 m 3 Solution 4: (a) Due to elastic collision, momentum and energy are conserved:, 2, 2 2 () (2) From () and (2) (b), After first collision between and, 0 and For the second collision, between and, we do not need to solve the equations again, we can write directly from (3) and (4) 2 Since, never returns back and collides with, so we have only two collisions. (c) Similarly, since, returns back and collides with. In this case, we have three collisions. 0 2

5 Question 5 [ Angular Motion ]: We could measure the speed of light, c, by using a rotating slotted wheel. As shown in the Figure 5, a beam of light passes through one of the slots at the outside edge of the wheel, travels to a mirror, and returns to the wheel just in time to pass through the next slot in the wheel. Assuming that the angular velocity of wheel is w, number of slots around wheel s edge is N, and the distance between the wheel and the mirror is L, find the speed of light c (20 points). Figure 5 Solution 5: In the time light takes to go from the wheel to the mirror and back again, the wheel turns through an angle The time is So the angular velocity of the wheel is 2 2

REVISING MECHANICS (LIVE) 30 JUNE 2015 Exam Questions

REVISING MECHANICS (LIVE) 30 JUNE 2015 Exam Questions Question 1 (Adapted from DBE November 2014, Question 2) Two blocks of masses 20 kg and 5 kg respectively are connected by a light inextensible string,