DHS 03 Dynamics (Tutorial Solutions)

DHS Year 5 Physics H2 Topic 3: Dynamics Tutorial Suggested Answers 2024 1 For Internal Use Only lim.boonsiong@dhs.edu.sg Topic 3: Dynamics Part 1: Newton’s Laws of Motion 1. A footballer accelerates a football horizontal from rest to a speed of 10 m s –1 during the time in which his toe is in contact with the ball for 0.20 s. If the football has a mass of 0.50 kg, what average force does the footballer exert on the ball? [25 N]  10 00.50 25 N0.20 vFm am t     2. The air exerts a forward force of 10 N on the propeller of a 0.20 kg model plane. If the plane accelerates forward at 2.0 ms –2, what is the magnitude of the resistive force exerted by air on the airplane? [9.6 N]     10 0.20 2.0 10 0.40 9.6 NFm a D D      3. A 5.0 g bullet leaves the muzzle of a rifle with a speed of 320 ms –1. What net force (assumed constant) is exerted on the bullet while it is moving down the 0.82 m long barrel? [312 N]     22 22 4 2 34 The bullet has constant acceleration down the barrel due to constant net force. 2 320 0 2 0.82 6.24 10 m s 5.0 10 6.24 10 312 N a vu a S a a Fm a             4. Two blocks, X and Y, of masses 3 kg and 2 kg respectively, are accelerated along a floor by a force 20 N applied to block X, as shown in the diagram. Given that the frictional forces that the floor acts on X and Y are 6 N and 4 N respectively, (a) Draw free body diagrams showing how the forces act on X and Y. (b) Base on Newton’s second law, write three equations that govern the motion of the objects (c) What are their acceleration(s)? [ 2 ms−2] (d) What is the magnitude of the force exerted by block Y on block X? [8 N] Contact forces Two solids are actually subjected to a pressure over the area where they are in contact. It is hard to calculate the pressure distribution, so instead we replace the pressure by a statically equivalent force. If the two contacting surfaces are flat, then (1) The reaction force can be modelled as a single force, with no moment (2) The force can act anywhere within the area of contact (its actual position is determined by force and moment balance) (3) The force must be perpendicular to the two surfaces (4) The force acts to repel the two solids. X 20 N Y

DHS Year 5 Physics H2 Topic 3: Dynamics Tutorial Suggested Answers 2024 2 For Internal Use Only lim.boonsiong@dhs.edu.sg (a) (b), (c) & (d)    C C -2 C C C Isolating both I Substitute value of into (1), X & Y, 20 6 4 3 2 Isolating Y, 4 2 (3) 10 2 m s5 sol we h ati ave ng X, 20 6 3 1 (2) 43 8 (1) 14 3 42 N F a a a Fa Fa Fa a Fa               ( FC is the force exerted by Y on X, F’C is the force exerted by X on Y. FC and F’C is a third-law pair.) X 20 N 6N FC mXg NX Y F’C 4N mYg NY NX is the normal reaction acted by the floor on X. NY is the normal reaction acted by the floor on X.

DHS Year 5 Physics H2 Topic 3: Dynamics Tutorial Suggested Answers 2024 3 For Internal Use Only lim.boonsiong@dhs.edu.sg 5. Three blocks are connected on a horizontal frictionless table and pulled to the right with a force T1 = 60 N. If m1 = 30 kg, m2 = 20 kg and m3 = 10 kg, find the tensions T2 and T3. [30 N, 10 N] Considering the whole system, i.e. consider m1, m2, and m3 as one object of mass m: Acceleration of system 260 1.0 m s60 Fa m   Taking rightward direction as positive, Isolating m1, T1 – T2 = m1a (1) = 30 (1.0) = 30 Isolating m2, T2 – T3 = m2a (2) = 20 (1.0) = 20 Isolating m3, T3 = m3a = 10 (1.0) = 10 N (3) Sub T3 = 10 N into (2) gives T2 = 30 N m = 60 kg T1 = 60 N a m1 T1 T2 a m2 T2 T3 a

DHS Year 5 Physics H2 Topic 3: Dynamics Tutorial Suggested Answers 2024 4 For Internal Use Only lim.boonsiong@dhs.edu.sg 6. Determine the force that an 80 kg man exerts on the floor of an elevator when it (a) is at rest; (b) rises with a constant velocity of 2.0 m s −1; (c) descends with a constant velocity of 2.0 m s −1; (d) rises with a constant acceleration of 2.0 m s −2; (e) descends with a constant acceleration of 2.0 m s −2 (f) accelerating downwards at 9.81 m s−2 [785 N, 785 N, 785 N, 945 N, 625 N, 0 N] w Isolating the man of mass and taking upward direction as positive, Newton's second law gives is the 'measured' weight of the man. It is also called 'apparent' weight or effective eight. m Nm N gm a (       1 1 2 0, 0, 0 80 9.81 785 N 2.0 m s , 0, 0 80 9.81 785 N 2.0 m s , 0, 0 80 9.81 785 N 2.0 m s , 2.0 9.81 2.0 80 11.81 945 N 2.0 va Nm gm Nm g va Nm gm Nm g va Nm gm Nm g a Nm gm Nm a                                a) (b) (c) (d) (e)       2 2 m s , 2.0 9.81 2.0 80 7.81 625 N 9.81 m s , 9.81 9.81 9.81 0 N Nm gm Nm a Nm gm Nm                 (f)

DHS Year 5 Physics H2 Topic 3: Dynamics Tutorial Suggested Answers 2024 5 For Internal Use Only lim.boonsiong@dhs.edu.sg 7. The figure shows a painter in a crate which hangs alongside a building. When the painter pulls on the rope, the force he exerts on the floor of the crate is 300 N. If the mass of the painter and the crate is 75 kg and 25 kg respectively, find the acceleration of the crate. [2.2 m s2 upwards] Considering forces on the crate, (1)T N mg ma   Considering forces on the painter, (2)T N Mg Ma   Subtracting equation (1) from equation (2),  2 2( )( ) 2 300 (75 25) 9.81 2.2 m s (upward)75 25 NM m gM m a a          Alternatively, we can treat the man and the crate as one system and draw its free body diagram. Therefore, 2 ... (3) The unknowns and can be solved by using any pair of the three eq TM m gM m a aT   uations (1), (2) and (3). positive 2T M+m (M+m)g a

DHS Year 5 Physics H2 Topic 3: Dynamics Tutorial Suggested Answers 2024 6 For Internal Use Only lim.boonsiong@dhs.edu.sg 8. Three crates of masses 3M, M and 5M are stacked on top of one another on the floor of a lift as shown on the right. When the lift is accelerating upwards with an acceleration a, the magnitude of the force mass M exerts on the mass 5M is given by A 4Mg – 5Ma B 4Mg – 4Ma C 4Mg + 4Ma D 4Mg + 5Ma Answer: C (There are a few ways to solve the problem.)   Let to be the normal force acted by the lift floor on the system of three crates. Apply Newton's 2nd law to the system of 3 crates: 35 Isolating 5 , Newton's 2nd la 3 w 5 99 : 5C N NM M M gM M M a N M NF MM M g g a           1 2 5 where is the (contact) force mass exerts on mass 5 55 9 9 Let be the contact force between mass 3 and mass 55 , and the contact O f or b 4 ce R 4 C C Ma F M M F N Mg Ma Ma Mg Mg Ma M FM M F aM g        11 21 21 Isolating 3 , Newton's 2nd law: 33 33 Isolating , Newton's 2nd la etween mass and 5 . Let be the contact force between mass and mas w