Mechanics Answers to Examples B (Momentum) - 1 David Apsley

TOPIC B: MOMENTUM ANSWERS SPRING 2019 (Full worked answers follow on later pages) Q1. (a) 2.26 m s 2 (b) 5.89 m s 2 Q2. 8.41 m s 2 and 4.20 m s 2 ; 841 N Q3. (a) 1.70 m s 1 (b) 1.86 s Q4. (a) 1 s (b) 1.5 s; 7.36 m s 1 ; 19.6 m s 2 Q5. (a) 456 N (b) 227 N Q6. 1.41 m s 2 Q7. (a) 0.24 Q8. 10.1 s Q9. (a) ; ; ; (b) (c) 17.2 m Q10. 3.54 m from the wall; speed 15.0 m s 1 Q11. (a) 24.7 m s 1 (b) 20.0 m Q12. (a) 0.764 m s 1 ; (b) (c) 0.779 Q13. (a) 10.2 m s 1 (b) (c) 0.693 Mechanics Answers to Examples B (Momentum) - 1 David Apsley

Q14. (a) 2.55 m s 1 (b) 0.631 (c) 45.1% Q15. Q16. 3.17 m s 1 at 40.9 to right of original direction; 4.16 m s 1 at 30 to left of original direction; percentage loss of energy = 24.0% Q17. (a) 35.6 m s 1 (b) 34.3º; 55.7º (c) 2.59 m Q18. (a) 7.00 m s 1 (b) (i) 9.81 m s 2 downward; (ii) 19.62 m s 2 upward (c) 4.20 m s 1 ; 2 kg (d) 2.40 m Q19. (across, up) = (0.275, 0.331) m, relative to the bottom left corner Q20. (across, up) = (7.31, 1.88) cm, relative to the bottom-left corner Q21. (a) T PA = 10 N; T QB = 20 N (b) 14.0 Q22. 1.62 m Q23. (a) topples (b) no movement Q24. (a) 1.2510 5 Pa (b) 2.38 m Q25. (a) 0.466 (b) 131 L Mechanics Answers to Examples B (Momentum) - 2 David Apsley

Question 1. (a) Let a be the acceleration of the 50 kg mass upward or the 80 kg mass downward. Let T be the tension in the rope. Resolving upward for the 50 kg mass: Resolving downward for the 80 kg mass: Adding, to eliminate T: Answer: 2.26 m s 2. (b) Resolving upward for the 50 kg mass: However, this time, T = 80g. Hence: Answer: 5.89 m s 2. Mechanics Answers to Examples B (Momentum) - 3 David Apsley

Question 2. Let x 1 be the distance moved to the right by the 100 kg mass and x 2 be the distance moved down by the 300 kg mass. Let T be the tension in the rope. Considering the amount of rope passing over the pulley from each side, Hence, There must be the same proportionality between accelerations: Resolving to the right for the 100 kg mass: (A) Resolving downward for the 300 kg mass: or (B) Eliminating T by 2(A) + (B): Hence, Then, Answer: accelerations of 100 kg and 300 kg masses are 8.41 m s 2, 4.20 m s 2 respectively; tension = 841 N. Mechanics Answers to Examples B (Momentum) - 4 David Apsley

Question 3. (a) Let x and v be the upward displacement and velocity of the weight. The net pull from the two sides of the cable on the weight is 2F. Hence, Integrating from v = 0 when t = 0, Putting t = 2, and noting that weight mg = 30N, so that, Answer: 1.70 m s 1. (b) From part (a), Integrating from x = 0 when t = 0: (*) This is not invertible analytically. Instead, solve numerically to find t when x = 2 m. There are many possible methods. Two such are given below. Method 1: (repeated trial) Denoting the RHS of (*) by f(t), we find x increases monotonically with t (as v is always positive) and so a solution to f(t) = 2 must lie between t = 1 and t = 2. Homing in on a solution by repeated trial gives t = 1.864 s Method 2: (iteration) When x = 2, one way of rearranging (*) is Mechanics Answers to Examples B (Momentum) - 5 David Apsley

This can be iterated (use the [ANS] button on your calculator) from, e.g., t = 1, to give t = 1.000, 1.651, 1.828, 1.859, 1.863, 1.864, 1.864,... Answer: 1.86 s. Mechanics Answers to Examples B (Momentum) - 6 David Apsley

Question 4. (a) The tension throughout the cable is F. The mass is subject to force 2F upward (F from each side of the loop of cable) and weight mg downward. Taking x, v and a positive upward (though you are quite at liberty to adopt positive downward if you prefer), force = mass acceleration gives (in kg-m-s units): When t is 0, v = 0 and the acceleration is negative, which is actually enough to show that the initial motion will be downward. However, to find for how long we need to integrate and find v explicitly. Integrating from the initial conditions (v = 0 when t = 0) to general (v, t): From its factors, v goes negative (i.e. downward motion) immediately after t = 0 and stays negative until t = 1. Note that downward motion continues until velocity v = 0, not the equilibrium point where a = 0 or forces are in balance. At the latter time v is actually maximal! Answer: downward motion until t = 1 s. (b) In terms of upward displacement x: Integrating: Hence, x = 0 when t = 0 (start time) and t = 3/2 s. At the latter time: Answer: returns after 1.5 s; speed = 7.36 m s 1 ; acceleration = 19.6 m s 2. Mechanics Answers to Examples B (Momentum) - 7 David Apsley

Question 5. m = 50 kg μ = 0.25 a = 2 m s 2 (a) Resolving perpendicular to the plane: Since the block is moving, friction is given by Resolving along the plane: Hence, Answer: 456 N. (b) As above for R and F. This time, resolving along the plane and noting that there are pulls of magnitude T (but different directions) from either side of the pulley: Hence, Answer: 227 N. Mechanics Answers to Examples B (Momentum) - 8 David Apsley

Question 6. m = 40 kg μ = 0.4 T = 100 N (a) Resolving perpendicular to the plane: The maximum friction is then: The motive force along the plane is This exceeds the maximum friction force. Hence, the block moves, friction is maximal (F = F max ) and the acceleration a is given by Answer: 1.41 m s 2. Mechanics Answers to Examples B (Momentum) - 9 David Apsley

Question 7. (a) At the limiting state (minimum coefficient of friction) the friction force F is maximal and given by F = μr and exactly balances the horizontal hydrostatic pressure force on the front wall of the dam. If L is the length of the dam in metres then the mass of the dam is Balancing forces vertically: Balancing forces horizontally: average pressure area = friction force If h is the depth of water then (the pressure at the centroid). Hence, Answer: minimum coefficient of friction = 0.24. (b) In reality, water will seep underneath the dam, producing upthrust and reducing R. The coefficient of friction would then have to be higher to provide the same friction force. It is useful to key the dam into the underlying bedrock, both to provide resistance to horizontal motion and reduce the seepage underneath it. Mechanics Answers to Examples B (Momentum) - 10 David Apsley

Question 8. Initial velocity: Using: impulse = change of momentum Answer: 10.1 s. Mechanics Answers to Examples B (Momentum) - 11 David Apsley

Question 9. (a) 1000 kg vehicle: 750 kg vehicle: (These can be rounded for a final answer). (b) Total momentum before = total momentum after Hence, which gives: (This can be rounded for a final answer). (c) Magnitude of velocity in part (b) is The acceleration is Using Answer: 17.2 m. Mechanics Answers to Examples B (Momentum) - 12 David Apsley

Question 10. Horizontally, the ball travels toward the wall at 20 m s 1 and travels back from it at Vertically, the ball s motion is not changed by the wall (it is the component parallel to the surface), so, since u y = 0 and a y = g, its vertical displacement from initial position is given for all t (i.e. before and after hitting the wall) by Setting y = 1.5 m gives time to first bounce on the ground: t = 0.5530 s. Time taken to reach wall at 20 m s 1 horizontally is The remaining time to the first bounce on the ground is 0.2530 s. The horizontal distance travelled back at 14 m s 1 is then Velocity and speed when it first hits the ground: Answer: 3.54 m from the wall; speed 15.0 m s 1. Mechanics Answers to Examples B (Momentum) - 13 David Apsley

Question 11. (a) Initial velocity: Displacement (as a function of t) by constant-acceleration formulae: Substitute in the latter displacement formula to get the trajectory: Substituting x = 25 m, y = 10 m, θ = 35, v 0 = 24.67 m s 1 Answer: 24.7 m s 1. (b) Time to hit the wall, The velocity components at this time are: Restitution affects only the velocity component perpendicular to the wall (i.e. the horizontal, x, component). Hence, the constant-acceleration motion in the second part of the motion has: The time for a vertical displacement of y = 10 m (from the height at which it hit the wall to when it hits the ground) is given by Mechanics Answers to Examples B (Momentum) - 14 David Apsley

(or a meaningless negative time) The horizontal position (relative to the wall) is then Answer: 20.0 m back from the wall. Mechanics Answers to Examples B (Momentum) - 15 David Apsley

Question 12. (a) Conservation of x-momentum: Answer: 0.764 m s 1. (b) Use: Impulse = change of momentum Answer:. (c) Fraction of energy lost: Answer: 0.779. Mechanics Answers to Examples B (Momentum) - 16 David Apsley

Question 13. Take the origin of coordinates at the initial point. Then x = 6 m and y = 2 m at C. Initial velocity: Displacement (as function of t) by constant-acceleration formulae: Substitute in the latter displacement formula to get the trajectory: Rearranging: Set x = 6 m and y = 2 m to get Answer: 10.2 m s 1. (b) The time taken is given by Then Answer: (c) After oblique collision, only the component of velocity perpendicular to the wall is changed. After collision: Mechanics Answers to Examples B (Momentum) - 17 David Apsley

Since we have Answer: 0.693. Mechanics Answers to Examples B (Momentum) - 18 David Apsley

Question 14. (a) The speeds of the driver before and after may be determined from energy (Topic C) or from the constant-acceleration formula. Taking positive direction downward: Before (u = 0, a = 9.81 m s 2, s = 2 m): velocity hitting pile = 6.264 m s 1. After (v = 0, a = 9.81 m s 2, s = 0.1 m): velocity leaving pile = 1.401 m s 1. By momentum conservation: Answer: 2.55 m s 1. (b) Coefficient of restitution: Answer: 0.631. (c) Percentage of energy lost Answer: 45.1%. Mechanics Answers to Examples B (Momentum) - 19 David Apsley

Question 15. Consider the collision between the (r 1) th and the r th sphere. The first of these has velocity v r 1 before collision and w r 1 after. Momentum: Restitution: (A) (B) Adding (A) + (B) to eliminate w r 1 : Hence, and so, inductively, Answer: Mechanics Answers to Examples B (Momentum) - 20 David Apsley

Question 16. By trigonometry (see the diagram) the line of centres when the balls collide obliquely is 30 from the original direction of motion. Adopt new x and y coordinates along the tangent to the surfaces at collision and along the lines of centres respectively. The second particle must move off in the direction along the line of centres (v 2x = 0). Note that u 1 = 6 m s 1. y 30 o r r x Momentum along tangent surface: r (A) u 1 Momentum along line of centres: (B) Restitution: (C) Adding (B) + (C) gives: (B) then gives In the new coordinate system, sphere 1 has velocity direction are, respectively:. The magnitude and - to right of line of centres (40.89 to original direction) In the new coordinate system, sphere 2 has velocity, with magnitude and is directed along the line of centres. Mechanics Answers to Examples B (Momentum) - 21 David Apsley

Percentage of energy lost: Answer: Sphere 1: 3.17 m s 1 at 40.9 to the right of the original direction; sphere 2: 4.16 m s 1 at 30 to left of original direction; percentage loss of energy = 24.0% Mechanics Answers to Examples B (Momentum) - 22 David Apsley

Question 17. (a) Impulse = change in momentum V = 35.56 m s 1 Answer: 35.6 m s 1. (b) Horizontal and vertical displacements from constant-acceleration formulae: Range (y = 0) corresponds to time and distance 2θ = 68.59º or 111.41º θ = 34.30º or 55.71º Answer: 34.3º or 55.7º. (c) Time to first bounce: Velocity components just before first bounce: Actually, the latter is obvious. By energy or symmetry, downward velocity on return is equal to initial upward velocity: (last digit different due to rounding). Velocity components after first bounce (oblique collision with e = 0.6): (parallel component unchanged) (perpendicular component subject to restitution) Mechanics Answers to Examples B (Momentum) - 23 David Apsley

Further time to hit the wall: Answer: 2.59 m. (d) v x(m/s) 29.38 Horizontal component: constant (no acceleration, and unaffected by the bounce) 0 t (s) v y(m/s) 20.03 12.02 0 bounce hits wall 4.085 6.297 t (s) Vertical component: constant downward slope (acceleration dv y /dt = 9.81 m s 2 ); reverses sign and magnitude changes by factor e = 0.6 at the bounce on the ground. -20.03 Mechanics Answers to Examples B (Momentum) - 24 David Apsley

Question 18. (a) Gain in KE = loss in PE Answer: 7.00 m s 1. (b) Particle A is moving in a circle, so: radial acceleration is v 2 /r and can be found directly from the speed v; tangential acceleration is dv/dt and can be found as tangential force/mass. (i) Initially, the speed v is zero, so acceleration is purely tangential: (tangential; i.e. downward here) (ii) At the lowest point there is no tangential force, so no tangential acceleration. The acceleration is purely centripetal (radially inward) of magnitude: (centripetal; i.e. upward here) Answer: (i) 9.81 m s 2 downward; (ii) 19.62 m s 2 upward. (c) Just before collision, particle A has speed u A = 7.004 m s 1 and particle B is at rest. Restitution: speed of separation = e speed of approach ) Conservation of momentum: Answer: speed = 4.20 m s 1 ; mass = 2 kg Mechanics Answers to Examples B (Momentum) - 25 David Apsley

(d) Particle B now undergoes two-component motion under gravity. Since all the subsequent motion is downward take x horizontal forwards and y positive downward. ; ; Find the time of impact: Horizontal distance travelled: Answer: horizontal distance travelled = 2.40 m. Mechanics Answers to Examples B (Momentum) - 26 David Apsley

Question 19. For a uniform plane lamina the centre of mass coincides with the centre of area. Measure coordinates x, y horizontally and vertically from (e.g.) the bottom-left corner. The frame can be obtained by subtracting rectangle 2 from rectangle 1, where the areas and centroids of the original rectangles are as follows. So that they can be used in a summation formula, subtracted areas are marked negative. Rectangle 1 (outer rectangle): a 1 = 0.5 0.7 = 0.35 m 2 x 1 = 0.25 m y 1 = 0.35 m Rectangle 2 (inner rectangle): Then, a 2 = 0.2 0.35 = 0.07 m 2 x 2 = 0.15 m y 2 = 0.425 m Answer: (across, up) = (0.275, 0.331) m, relative to the bottom left corner. Mechanics Answers to Examples B (Momentum) - 27 David Apsley

Question 20. For a uniform plane lamina the centre of mass coincides with the centre of area. Divide the shape into three parts (triangle / rectangle / triangle) whose areas and individual centres of mass / centres of area are: Then: Answer: (across, up) = (7.31, 1.88) cm, relative to the bottom-left corner. Mechanics Answers to Examples B (Momentum) - 28 David Apsley

Question 21. (a) The centroid of a triangle lies at 1/3 rd of the altitude from any side; here it is 50 mm from AB and 100 mm from BC. The weight of 30 N acts through this point. Taking moments about B: Taking moments about A: Answer: T PA = 10 N; T QB = 20 N. (b) Answer: 14.0. Mechanics Answers to Examples B (Momentum) - 29 David Apsley

Question 22. For a uniform plane lamina the centre of mass coincides with the centre of area. Take x and y from the corner at which the lamina is suspended, with y along the long side. By symmetry,. The shape can be formed by subtraction of a circle from a rectangle. Working in metres throughout we have the following. So that they can be used in a summation formula, subtracted areas are marked negative. Part 1 (rectangle): Part 2 (circle): Then But, from the angle of suspension, whence Hence, Solving a quadratic equation: Since L is the longer side, here it takes the value 1.618 m. Answer: 1.62 m. Mechanics Answers to Examples B (Momentum) - 30 David Apsley

Question 23. One must consider, in this order: toppling, if the vertical through the centre of mass lies outside the base; sliding, if the downslope component of weight is greater than maximum friction. (a) The centre of mass is at height which this acts on the slope is. The distance from the cylinder axis at This is greater than the base radius 0.05 m, so that the line of action of the weight lies outside the base of the cylinder. The weight then produces an unbalanced moment about the bottom of the cylinder, so the cylinder will topple over. Answer: topples. (b) The centre of mass is at height axis at which this acts on the slope is. The distance from the cylinder This lies within the confines of the base of the cylinder and hence the cylinder does not topple. The maximum friction force is The component of weight downslope is The ratio of these two along-slope forces is so that friction is sufficient to prevent motion. Answer: no movement. Mechanics Answers to Examples B (Momentum) - 31 David Apsley

Question 24. (a) (*) where ρ is density, A is frontal area, g is gravity, W is width of one column. Note that the unknown depth D cancels, because both the weight and the base area are proportional to it. Divide the front up into two equal-area rectangles (1 and 2), outer semicircle (3) and inner semicircle (4), where the last is to be subtracted. Working in metres throughout we have the following. Subtracted areas are marked negative. Total area (in m 2 ): From (*), the average pressure exerted on the foundations is Answer: average pressure = 1.2510 5 Pa. (b) Height of the centre of mass: Answer: height of centre of mass = 2.38 m. Mechanics Answers to Examples B (Momentum) - 32 David Apsley

Question 25. (a) If R is the normal reaction, F max the maximum friction force, m is the mass of container and water and θ = 25º is angle of slope then, resolving perpendicular to the plane: There is no slipping if the motive force (here, the downward component of weight) is less than maximum friction: i.e. Answer: 0.466. (b) The volume of water in the container, when seen in cross-section, forms either a triangle with hypotenuse L or, once there is sufficient water, a rectangle of height h, width L, plus triangle (see diagram). Taking upslope coordinate x and perpendicular coordinate y relative to the point about which the container will topple, then toppling will occur once the centre of mass is forward of the vertical line through this point; i.e. if (*) h tan 25 o 25 o h 2 y 1 x L=0.4 m From the geometry of the figure, with 1 indicating the rectangle (if there is sufficient water so that h > 0): Then, and so (*) gives toppling if Mechanics Answers to Examples B (Momentum) - 33 David Apsley

(**) For the case h = 0 (i.e. before the base of the vase is completely covered) this would require which does not occur when θ = 25º (or, indeed, for any other angle below 45º). Otherwise, we have L = 0.4 m (the side of the container) and (**) gives The only positive root of the quadratic is, by formula, 0.7286 m. The centre of mass will be forward of the toppling line once this is exceeded. The volume of water is then Answer: 131 L. Mechanics Answers to Examples B (Momentum) - 34 David Apsley