HCI 02 Kinematics Solutions (Discussion Questions)

1 Tutorial 2A: Kinematics Discussion Questions (Suggested Solution) D1 (a) Yes. An object in uniform circular motion moves at constant speed but its direction of motion is changing all the time (its velocity is changing) and hence it has acceleration. Circular Motion will be covered in greater detail in chapter 6. (b) No. The ma gnitude of the velocity is the speed of an object. If the velocity is constant, both magnitude and direction of velocity have to be constant. Hence, speed cannot change. (c) Yes. Projectile motion of a projectile under free -fall, with no air resistance . The direction of motion (velocity) changes with time. The acceleration, g, is constant. (d) Yes. The object can be instantaneously at rest and the next moment its velocity increases or decrease. Eg an object thrown vertically upwards and at its highest p oint, it is instantaneously at rest but it is still accelerating downwards with g. (e) Yes. That happens when the object is slowing down. (f) No. Consider an object initially moving with some vel ocity then resting for some time then continuing to move with some velocity. The average velocity is not zero as net displacement is not zero but during this interval it was at rest at some point. D2 At max height, velocity is zero and gradient at that point is equal to acceleration of free fall. Also, point A is the point it first hits the ground. NOTE: area of triangle ABC should be equal size to area of triangle CD since the ball rises and falls through the same distance after the first bounce. Ans: C D3 The area under velocity-time graph gives the displacement. Both objects have the similar acceleration rates throughout. You may sketch the displacement versus time graph for both objects. The displacement for P is obviously much greater. In fact, the area under velocity-time for Q suggests that the displacement of Q is zero at the end of time t2. To compare the distance travelled by P and Q, we could make use of the area under velocity-time graphs Object P: 2 2 vt

2 Object Q: ( )211211 222 2 2 2 2 2 4 ttt vtvv  −  +  =   . The distance travelled is also different. Ans: A D4 The area under velocity-time graph gives the displacement of the cars. At t = 0, both are at the same position where Car X takes over. At t = T, both cars must also be at the same position so that Car Y could take over. The displacement of Car X at t = T is P + Q + R and that of car Y is S + Q + R. The displacements must be the same. Hence, P is equal to S. Ans: A D5 Area under the graph represents change in velocity. At point C, the area (from start to point C) is greatest. Beyond point C there is a negative change in velocity, which means the velocity is decreasing. Ans: C D6 Method I (equations of motion) Let t = 0 be when the ball passes by light gate 1. At light gate 2, 2 1 1 2s ut at=+ (1) At light gate 3, 2 12 1(2 ) (2 )2s s u t a t+ = + (2) (2) - 2×(1): 22 21 21 2 1 (4 2 )2s s a t t ssa t − = − −= Ans: A Method II (graphical) Assume the speed of the ball is u when it passes the light gate 1 at time t = 0 and accelerates constantly with a. Sketching v-t graph,

3 The area under the v-t graph is displacement, hence ( ) ( ) ( ) ( ) ( ) 2 1 2 2 22 21 21 2 11 ( 0) 222 11 2 (2 ) 2 322 1 32 S u u at t ut at S u at u at t t ut at SSS S at at a t = + + − = + = + + + − = + −− = − → = D7 Stage 1 => sphere in air, with negligible air resistance and upthrust => free-fall, a = g => option A & E not possible Stage 2 => sphere enters fluid, there is drag force. => 2 possible cases➔ Case 1: mgFD  , Case 2: mgFD Case 1: mgFD  Sphere slows down as a is negative, reaches terminal velocity when a = 0 m s-2 Case 2: mgFD Sphere continues to accelerate downwards but at a < g. Velocity increases, DF increases until mgFD = . Net force on sphere is zero and a = 0. Sphere reaches terminal velocity. (option not available) Ans: B t 2t t v u u + at u + 2at S1 S2 0 Stage 2 FD W W Stage 1

4 D8 Option D & E are not possible as H does not change after some time, implying that ball bearing stays stationary which is not possible. Since ball is released, its initial velocity is zero and hence gradient of h-t graph at t = 0 s should be zero. At t > 0 s, as ball bearing speeds up, FD increases. m Fmga D−= , ga , ball bearing continues to speed up at slower rate. Eventually, mgFD = , net force acting on ball bearing = 0, a = 0 and ball bearing reaches terminal velocity (speed constant, gradient constant). Ans: A D9 (a)(i) 20 m s-1 (ii) =Aa gradient of v-t graph 010 020 − −= 0.2= m s-2 (iii) =Ea gradient of v-t graph 5055 30)5( − −−= 0.7−= m s-2 (iv) =Bs area in section B 1520= 300= m (v) =Cs area of trapezium in C 10)3020(21 += 250= m (b) At t = 50 s, the object decelerates uniformly at a rate of 0.7 m s-2 from 30 m s-1 until it comes to an instantaneous rest at 54t s. It then moves back (in opposite direction to its original motion) and accelerates at 0.7 m s-2 until it reaches a speed of 5 m s-1 at t = 55 s. It continues at a uniform speed of 5 m s-1 in the negative direction for the next 10 s until t = 65 s. (c)

5 Note: Sections B, D and F are straight. The peak occurs at about 54.5 s. D10 (a) There is air resistance, which increases with the speed of the object. At a large enough speed, the air resistance equals the weight of the object. Net force becomes zero since weight and air resistance act in opposite directions, hence acceleration becomes zero and velocity becomes constant. (b) (i) By Newton’s 2nd law, netF mg kv=− ( ) / ma mg kv g a kv m =− −= (ii) At v = 0 m s-1, a = 9.81 m s-2 because the object experiences no air resistance when it is not moving. At v = 40 m s-1, a = 0 m s-2 because the object is at terminal velocity and net force is zero. The acceleration at v = 30 m s-1 is calculated from the gradient of the tangent at v = 30 m s-1. (iii) If the student’s suggestion is correct, then ga v − is constant. At v = 20 m s-1, 1.6 0.08020 ga v − == At v = 30 m s-1, 4.1 0.13730 ga v − ==

6 The student cannot be correct since the percentage difference between the values is large. [ 0.137 0.080 100% 53%0.137 0.080 2 − =+  ] D11 (c) (i) At maximum height, the ball is instantaneously at rest. From the graph, v = 0 m s-1 at t = 1.80 s. (ii) The acceleration at an instant can be determined from the gradient of the tangent line at that instant. (iii)It could be because there is significant air resistance acing on the ball. Air resistance is present as the ball possesses velocity at the point which it is thrown. The air resistance is acting in the opposite direction as its velocity, which is in the same direction as its weight. Hence the downward acceleration is greater than the gravitational pull of g. (iv)At t = 1.80 s, the ball is at rest and air resistance is zero, hence only its weights acts on it. It experiences an acceleration of g. (v) (d) On its flight up, the drag force acting on the ball is in the same direction as its weight. On its flight down, the drag force acting on the ball is in opposite direction to its weight. Hence, the average acceleration on its flight up aAB is greater than the average acceleration on its flight down aBC. Using asuv 222 =− , From A to B, sav ABA 20 22 =− sav ABA 2 2 −= ----- (1) From B to C, sav BCc 2022 =− sav BCc 2 2 = ----- (2) Since aBC < aAB, comparing (2) and (1), the final speed reached by the ball when it falls back onto the hand vc is smaller than the initial speed of projection vA. Hence, the average