HCI 02 Kinematics Solutions (Self-Review Questions)

1 Tutorial 2A: Kinematics Self-Review Questions (Suggested Solution) S1 (a) Change in speed = final speed – initial speed = (30 – 10) m s-1 = 20 m s-1 (b) Change in velocity Change in velocity = vfinal – vinitial = vfinal + (– vinitial) Magnitude of the change in velocity Using cosine rule, v = √(102 + 302 – 2(10)(30) cos 60o) = 26.5 m s-1 Direction of the change in velocity Using sine rule, sin α sin 60o 10 = 26.5 α = 19.1o The change in the velocity is 26.5 m s-1 at an angle of 79.1o anticlockwise from the horizontal (c) Average speed = total distance travelled total time taken = (10 x 5.0) + (30 x 2.0) (5.0 + 2.0) = 15.7 m s-1 (d) Average velocity Using cosine rule, ∆𝑠⃗ = √(602 + 502 – 2(60)(50) cos 120o) = 95.4 m Magnitude of the average velocity, |v|= 95.4 / 7.0 = 13.6 m s-1 Using sine rule, β = 33.0o The average velocity is 13.6 m s-1 at 33.0o to the initial velocity. S2 (a)To find total displacement AC, =ABs 6.4960 3585 = km Using cosine rule, )4590cos(2222 oo BCACBCABAC sssss +−+= )135cos(1306.4921306.49 222 o ACs −+= 169=ACs km Using sine rule, N A C 35 min, 85 km h-1 2 h, 130 km Stop for 15min B 

2 o ACBC ss 135sinsin = o0.33= (b)Average velocity = 6.59 260 15 60 35 169 = ++ km h-1, o033. east of north Points to note: Students are expected to be able to recall the cosine and sine rule as these formulas are not given in any Physics test/exam. Answers should only be written in terms of North, South, East, West if such terminology were used in the question in the first place. S3 (a)Time taken = Distance travelled / speed = 14304 04 .. ).( == v R s (b) average velocity total change in displacement total time= = 552143 0404 .. .. =+ m s-1 A to B (c) change in velocity, 0.8=−= if vvv  m s-1, upwards At A, iv At B, fv For v S4 (a)average velocity between t = 0.0s and 3.0s, total change in displacement total time= = 83.00.00.3 0.45.1 −=− − m s-1 (b) At t = 3.0s, instantaneous velocity = 𝑑𝑥 𝑑𝑡 = gradient of tangent line at t = 3.0s = 67.02.50.0 0.05.3 −=− − m s-1 Exam skills: The 2 points chosen as coordinates for the gradient of the tangent line must be at least ½ the span of the graph. Points to note: A common misconception is students has the wrong concept of velocity = 𝑥 𝑡 and think that instantaneous velocity = 𝑥 𝑡 = 1.5 3 = 0.5 m s-1 . S5 Gradient of d-t graph is v-t graph. Note that at t = 0, the v is zero and hence the gradient of the d-t graph at t = 0 must be flat. When the object reaches terminal velocity (v-t graph flattens out horizontally), the gradient of the d-t graph becomes constant (non-zero). Ans: C S6 Displacement = area under velocity-time graph = (area from 0 s to 5 s) – (area from 5 s to 7 s) = 22212521 − = 3 m Ans: B Possible extensions: Can you visualise the motion of the particle and sketch a graph to show its motion?

3 S7 For acceleration to have the greatest numerical value, the change in velocity has to be the greatest  Slope of the graph has to change the most In fact, for all the other options the acceleration is zero. Ans: B S8 (1) Slope of s-t graph gives the v-t graph (2) v-t graph is a continuous graph => no kinks in s-t graph Ans: C S9 (a) Consider the package at the point of released, u=5.0 m s-1 ( ) 2 21 atuts += 2)81.9(210.50.21 tt −+=− 62.1−=t s (NA) or 64.2=t s (b) ( ) atuv += )64.2)(81.9(0.5 −+=v 9.20−=v m s-1 (negative means downward direction) Points to note: The sign convention that you chose is important in kinematics. Do indicate the sign convention that you are taking so that your working is clear to the examiner. Note that even if you take a different direction for part (b), that is, the sign convention chosen is downwards as positive, (↓) 𝑣 = 𝑢 + 𝑎𝑡 𝑣 = −5.0 + (9.81)(2.64) = 20.9 m s-1 (downwards) You will still obtain the same answer! u=5.0m s-1, t=0s, s=0m 21.0 m

4 Tutorial 2B: Kinematics Self-Review Questions (Suggested Solution) S1 Parabolic trajectory => projectile motion in the absence of air resistance (a) speed: not constant (b) acceleration: constant (c) horizontal component velocity: constant (d) vertical component velocity: not constant (due to gravitational pull) S2 (a) Using ( ) 2 21 tatus yyy += 2)81.9(210300 t+= 82.7=t s (b) To find v, we will need vx and vy. ( ) 50===→ uuv xx m s-1 ( ) yyyy sauv 2 22 += )300)(81.9(20 2 +=yv 7.76=yv m s-1 222 yx vvv += 6.917.7650 22 =+=v m s-1 50 7.76tan = o0.57= below horizontal Points to note: Velocity is a vector quantity, hence the direction must be given, if not marks will be deducted. A good practice is to draw the vector diagram, indicate the angle 𝜃 in the diagram, and write a short description of the direction. (c) ( )→ 2 21 tatus xxx += 39182.750 ==xs m S3 Using ( ) 2 21 tatus yyy += 2)81.9(21)70sin900(1700 tto −+= 170=t s or 03.2=t s Ans: A S4 To find v, we will need vx and vy. ( ) 40===→ uuv xx m s-1 ( ) tauv yyy += 4.29)3(81.90 =+=yv m s-1 504.2940 22 =+=v m s-1 Ans: C  x  u=ux=50 m s-1 a=g vx vy u=900 m/s 1700 m u=ux=40 m/s

5 S5 ( )→ 2 21 tatus xxx += 960.424 ==xs m ( ) 2 21 tatus yyy += msy 780.481.9210 2 =+= Ans: A S6 ( )→ 2 21 tatus xxx += )1(00.1 −−−−= tux ( ) 2 21 tatus yyy += 281.921052.0 t+= 326.0=t s -------(2) Subs. (2) into (1), 07.3326.0 00.1 ==xu m s-1 S7 ( )→ xxs u t= )1()cos40(150 −−−−= t Let t be the time taken for the particle to reach 150 m horizontal range. ( ) 2 21 tatus yyy += )2/(sin0 tgu −=  )2(sin)40(2 −−−−= gt  Subst. (2) into (1),       = g  sin402)cos40(150 81.9 2sin40150 2 = = o4.33= or o6.56=

6 S8 Let the initial speed be u. ( )→ sx = ux t 100 = (u cos  )(2.0) u cos  = 50 -------------- (1) ( ) 2 21 tatus yyy += 0 = (u sin  )(2.0) + (0.5)(- 9.81)(2.0)2 u sin  = 9.81 ------------- (2) Taking (2) / (1), tan  = 9.81/50  = 11⁰