H203 Dynamics - 2. Tutorial Solutions (1718)

H203 Tutorial Solutions- Dynamics Page 1 of 8 Self Attempt Questions 1. (a) The time of contact is prolonged while the change in momentum remains the same. Hence the average force applied on the ball by hand will be reduced . By Newton’s 3rd law, the force on hand by ball will be reduced. (b) Prolonging the time of contact, it allows the force to be applied onto the ball for a longer time thereby increasing the change in momentum of the ball. (c) By Newton’s 1 st law, the o bject should fly forward when bus braked suddenly. The claim is not valid. 2. No. It only means that the net force is zero. For example, there can be two forces of equal magnitude acting on the object in opposite direction. 3. (a) Not possible. The acceleration must be in the same direction as the net force in accordance with Newton’s Second Law. (b) Possible. Example: An object thrown vertically upwards. 4. By Newton’s 2nd law, Net force, F = ma 10 – FD = 0.20 x 2.0 FD = 10 – 0.40 = 9.6 N 5. Initial and final velocities of bullet u = 0 ms-1 and v = 320 ms-1 respectively. Since the force on bullet is assumed constant, the bullet undergoes constant acceleration a where a may be found from kinematics: v2 = u2 + 2as 4 2222 1024.6)82.0(2 0320 2  s uva ms-2 By N2L : force on bullet, F = ma = 5.0x10-3 x 6.24 x 104 = 312 N a = 2.0 ms–2 10 N FD = ?

6. 𝐹𝑓𝑜𝑜𝑡𝑏𝑎𝑙𝑙𝑒𝑟 𝑒𝑥𝑒𝑟𝑡𝑠 𝑜𝑛 𝑏𝑎𝑙𝑙 = 𝑑𝑃𝑏𝑎𝑙𝑙 𝑑𝑡 ≈ 𝑃𝑓 − 𝑃𝑖 ∆𝑡 = 𝑚(𝑣𝑓 − 𝑣𝑖) ∆𝑡 = 0.50(10 − 0) 0.20 = 25 N acting in the direction of the initial motion of the ball. 7. (a) Impulse = change in momentum = m(vf – vi) = (70.0)(5.20 – 0) = 364 Ns (b) 𝐹𝑜𝑛 𝑝𝑎𝑠𝑠𝑒𝑛𝑔𝑒𝑟 = 𝑑𝑃𝑝𝑎𝑠𝑠𝑒𝑛𝑔𝑒𝑟 𝑑𝑡 ≈ ∆𝑃 ∆𝑡 = 364 0.832 = 438 N 8. No. The initial total momentum of the system is not zero. For both objects to be at rest, the final total momentum of the system is zero. This would not be possible as it would violate the Principle of Conservation of Momentum. The assumption is there is no net unbalance external force. 9. Initially the clay has momentum directed towards the wall. When it collides and sticks to the wall, it appears that the momentum is zero. It is tempting to conclude that momentum is not conserved. In reality, the momentum of the clay is transferred to the wall and Earth, causing both to move although the speed is too small to be observable due to the enormous mass of wall and Earth. 10 Linear momentum is not conserved . As the ball rolls down an incline, it experiences resultant force acting on it. The impulse will result in a change in momentum hence the momentum will increase. Linear momentum is only conserved when there is no external resultant force acting on the ball.

Tutorial Discussion Questions 11. (a) FYX: force by Y on X WX: weight by Earth on X RX: normal reaction force by floor on X fX: frictional force by floor on X FXY: force by X on Y WY: weight by Earth on Y RY: normal reaction force by floor on Y fY: frictional force by floor on Y (b) Consider forces on X: Fa – FYX – fX = mXaX 20 – FYX – 6 = 3a (Note: aX = aY = a) 14 – FYX = 3a ………..(1) Consider forces on Y : FXY – fY = mYaY FXY – 4 = 2a ................ (2) Consider forces on (X and Y) : Fa – fX – fY = (mX+mY) a 20 – 6 – 4 = 5a 10 = 5a …………..(3) (c) From equation (3), a = 2 ms-2 (d) From equation (2), FXY – 4 = 2 FXY = 8 N By N3L, magnitude of force by Y on X, FYX = FXY, magnitude of force by X on Y FYX = 8 N Y FXY WY RY fY X 20 N FYX Wx fX Rx

12. (a) When the acceleration is 2.0 ms -2, what is the force exerted by the tow -bar on the trailer? [3000N] Consider free-body diagram of trailer. NmafT mafT maFnet 3000   (b) When the tractor and the trailer are moving at a constant speed of 6.0 ms -1, what is the force exerted on the tow-bar by the trailer? [1000 N] Moving at a constant speed of 6.0 ms-1 means no acceleration. Force by tow-bar on trailer = Frictional force = 1000 N By Newton’s 3rd Law, force by trailer on tow-bar = force by tow-bar on trailer = 1000 N 13. Using the principle of conservation of momentum and taking vectors in the direction of the motion of the man to be positive: initial momentum = final momentum 0 = 𝑚𝑚𝑎𝑛𝑣𝑚𝑎𝑛 + 𝑚𝑏𝑜𝑜𝑘(−𝑣𝑏𝑜𝑜𝑘) = (730 9.81) 𝑣𝑚𝑎𝑛 + 1.2(−5.0) ∴ 𝑣𝑚𝑎𝑛 = 0.0806 m s−1 Time taken for man to travel reach the shore = 5.0 0.0806 = 62 s T f = 1000 N

14. (a) Initial weight of the rocket W = mg = (1.90 x 103) x 9.81 = 18639 N (b) By Newton’s 2nd Law: 𝐹𝑟𝑜𝑐𝑘𝑒𝑡 𝑒𝑥𝑒𝑟𝑡𝑠 𝑜𝑛 𝑔𝑎𝑠 = 𝑑𝑃𝑔𝑎𝑠 𝑑𝑡 ≈ 𝑃𝑓 − 𝑃𝑖 ∆𝑡 = 𝑚(𝑣𝑓 − 𝑣𝑖) ∆𝑡 = 7.40 (2.50 × 103) 1 = 18500 N downwards By Newton’s 3rd Law, the thrust (the force which the gas exerts on the rocket) = 18500 N upwards On ignition, the thrust acting upwards is less than the weight of the rocket. After some time, due to the ejection of gas from the rocket, the weight decreases such that the thrust is larger than the weight. The rocket is only then able to take-off. Weight of fuel needed to be burn before take-off = 18639 – 18500 = 139 N Mass of fuel needed to be burn before take-off = 139 / 9.81 = 14.2 kg Time needed before take-off, t = 14.2 / 7.4000 = 1.9 s 15. (a) Impulse = area under F-t graph = 8)2)(53(2 1  Ns (b) Impulse = change of momentum = mvf - mvi Final velocity vf = (impulse + mvi)/m = [8 + 1.5(0)]/1.5 = 5.3 ms-1. (c) Final velocity vf = (impulse + mvi)/m = [8 + 1.5(-2.00)]/1.5 = 3.33 ms-1. 16. 𝑑𝑃𝑏𝑎𝑙𝑙 𝑑𝑡 = 𝐹𝑛𝑒𝑡 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑡ℎ𝑒 𝑏𝑎𝑙𝑙 = W𝑏𝑎𝑙𝑙 = 2 × 9.81 = 19.6 N

17. In 1 hr, the mass of water hitting the roof = 𝑣𝑜𝑙𝑢𝑚𝑒 × 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = (𝐴𝑟𝑜𝑜𝑓 × 0.040)(1000) = 40𝐴𝑟𝑜𝑜𝑓 𝑘𝑔 Taking upward vectors to be positive, the force which the roof exerts on the water = ∆𝑃𝑤𝑎𝑡𝑒𝑟 𝑡 = 40𝐴𝑟𝑜𝑜𝑓[0 − (−10)] 60 × 60 = 0.111𝐴𝑟𝑜𝑜𝑓 N (𝑢𝑝𝑤𝑎𝑟𝑑𝑠) By Newton’s 3rd Law, force which the water exerts on the roof = 0.111𝐴𝑟𝑜𝑜𝑓 N (𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑𝑠) Hence pressure = 𝐹𝑜𝑟𝑐𝑒 𝐴𝑟𝑜𝑜𝑓 = 0.111 𝑃𝑎 18. (a) The principle of conservation of linear momentum states that the total momentum of a closed system of colliding objects remains constant if no unbalance external force act on the system. (b) For elastic collision, both the total linear momentum and the total kinetic energy of the colliding objects are conserved. For inelastic collision, the total linear momentum of the colliding objects is conserved but the total kinetic energy is not conserved. (c) (i) Take vectors to the right as positive. Let the velocity of the 5.00 kg mass be v1 and that of the 10.0 kg mass to be v2. Using conservation of momentum, initial total momentum = final total momentum (5.00) (20.0) = 5.00 v1 + 10.0 v2 . . . . (1) Elastic collision means that total kinetic energy is conserved, 2 2 2 1 2 )0.10(2 1)00.5(2 1)0.20)(00.5(2 1 vv  . . . . (2) (Alternatively can use vrelative of approach = vrelative of separation) Solving (1) and (2): v1 = -6.67 ms-1 v2 = 13.3 ms-1 (ii) Fraction of initial KE transferred to the 10.0 kg object = u v i f KE KE 2 1 2 2 1 2 )00.5(2 1 )0.10(2 1 )( )(  = 2 ( )0.20 3.13 2 2 = 0.884

19. (a) (i) The linear momentum of a body is the product of its mass and its velocity. (ii) The change in linear