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Ls 11 Components of vectors
- 1. KYS Malaysia Physics AS / 2 S.M.Chappell Lesson 11 2013-2014 際際滷 1 of 12 Components of Vectors.
- 2. KYS Malaysia Physics AS / 2 S.M.Chappell Lesson 11 2013-2014 際際滷 2 of 12 Starter Solve the following problems Q1. A ship is pulled by two tugs at 12o to the direction the ship. If each pulls with a force of 6kN. Calculate the resultant force the ship is moved with Q2. A kite has a mass of 0.5 kg. It is hit by a find with a force of 10N. (Assume g = -9.81 ms-2) 10 N a) Find the weight of the Kite. b) Using your answer to a) construct a triangle for the forces c) Using your answers to a) and b) find the tension in the main kite string
- 3. KYS Malaysia Physics AS / 2 S.M.Chappell Lesson 11 2013-2014 際際滷 3 of 12 Objectives 1) Know the effect of gravity on the movement of . a projectile in terms of components 2) Understand how to resolve projectile problems through resolution of vectors into components 3) Be able to solve projectile problems
- 4. KYS Malaysia Physics AS / 2 S.M.Chappell Lesson 11 2013-2014 際際滷 4 of 12 Keywords Acceleration Component Constant Cosine rule Displacement Flight Path Horizontal Component Magnitude Maximum Pythagoras theorem Speed Sine rule Trajectory Varies Vectors Velocity Vertical Component vutas equations
- 5. KYS Malaysia Physics AS / 2 S.M.Chappell Lesson 11 2013-2014 際際滷 5 of 12 Objects thrown upwards. In these sort of problems it is convention that upwards is a positive direction and downwards a negative direction Look at the following problem Q1 .A person throws a cricket ball straight up into the air with an initial velocity of 30ms-1. What height will the ball reach from its point of release? (Assume g= -9.81ms -1) Draw a diagram to represent the information 30ms -1 Height 9.81 ms -2 Assign the correct signs to the vectors - Q. At what point will the ball stop increasing in height? When the upward (+) instantaneous velocity is zero Summarise further information now u = 30 ms-1 v = 0 ms-1 a = -9.81 ms-2 Height = displacement (s) s = ? Looking at the data use the appropriate vutas equation v 2 = u2 + 2as v = u +at Eqn 1 s = (v + u)t Eqn 2 2 s = ut + 遜 at2 Eqn 3 v2 = 2as + u2 Eqn 4 v 2- u2 = s 2a 0 2- 302 = s 2 -9.81 - 900 = s -19.62 s = 45.8 m
- 6. KYS Malaysia Physics AS / 2 S.M.Chappell Lesson 11 2013-2014 際際滷 6 of 12 Objects thrown upwards. In these sort of problems it is convention that upwards is a positive direction and downwards a negative direction Look at the following problem Q2 .A person throws a ball straight up into the air with an initial velocity of 27ms-1, and it goes over a cliff 40m high. How long does it take to reach the top of its trajectory? What is the final velocity the ball will reach the ground at the bottom of the cliff? Draw a diagram to represent the information 27 ms -1 Height 9.81 ms -2 Assign the correct signs to the vectors - u = 27 ms-1 v = ms-1 a = -9.81 ms-2 Height = displacement (s) s = - 40 m v = u +at Eqn 1 s = (v + u)t Eqn 2 2 s = ut + 遜 at2 Eqn 3 v2 = 2as + u2 Eqn 4 40 m- v = u +at Eqn 1 0 = 27 +(-9.81 x t ) 0 -27 = -9.81 x t 0 -27 = t -9.81 t = 2.75 s Find the time for the object to reach the top of its height v2 = 2as + u2 Eqn 4 Final velocity from full drop v2 = 2 x -9.81 x -40. + 272 v2 = 1513.8 v2 = 1513.8.8 v = - 38.91 ms-1
- 7. KYS Malaysia Physics AS / 2 S.M.Chappell Lesson 11 2013-2014 際際滷 7 of 12 Components of vectors. Just as several vectors can be made to give a resultant vector the opposite is true. In this that a vector can be split into two perpendicular component vectors. This becomes useful, particularly for projectiles. Imagine the problem you have just solved, except the ball was thrown horizontal. With initial velocity of 30ms-1. Q2. What would the flight path of the ball be? Down Curved Q3. Why will it follow an curved down path? Acceleration due to gravity Constant velocity 30ms-1 (Assuming no drag which is arrows same length common for these type of questions) Velocity at each point due to the acceleration due to gravity increases Longer downward arrows So there is still a problem. The velocity varies vertically
- 8. KYS Malaysia Physics AS / 2 S.M.Chappell Lesson 11 2013-2014 際際滷 8 of 12 Acceleration due to gravity Timeinmilliseconds Look at the object falling opposite. At each millisecond the position of an identical ball is shown and a line drawn to the axis Q4. What do you notice about the distance the ball moves in the y (downwards) direction ? Moves the same amount downwards or negative direction This enables us along with the use of vutas equations to solve these types of problems v =u + at Equation 1. s = (v + u ) t Equation 2. 2s = u t + 遜 at 2 Equation 3. v2 = 2as + u2 Equation 4. Plus the use of the trigonometric formulas a = b = c . sin a sin b sin c Sine rule Pythagorus theoremCosine rule a2 = b2 + c2 2bc cos A
- 9. KYS Malaysia Physics AS / 2 S.M.Chappell Lesson 11 2013-2014 Solving projectile problems 際際滷 9 of 12 Consider the following problem: A cannon fires a steel ball at 83 ms- 1 in the horizontal direction from the top of a cliff 80m high. Find the maximum range of the canon. (g = -9.81ms-2) Draw out the problem 80m s Solve vertical function. The time the cannonball will stay in the air s = u t + 遜 at 2 Equation 3. Initial velocity u is zero s = 0 x t + 遜 at 2 -80m = 遜 -9.81 t 2 -80m x 2 = t 2 -9.81 160/9.81 = t2 16.31 = t2 4.04 s = t s = (v + u ) t Equation 2. 2 Now the horizontal component s = (83 + 83 ) 4.04s 2 s = (166 ) 4.04s 2 s = 83 x 4.04s s = 335.32 m v =u + at Equation 1. s = (v + u ) t Equation 2. 2 s = u t + 遜 at 2 Equation 3. v2 = 2as + u2 Equation 4. Sine rule a = b = c . sin a sin b sin c Cosine rule a2 = b2 + c2 2bc cos A Pythagorus theorem Or s = v t s = 83 x 4.04 = 335.32 m
- 10. KYS Malaysia Physics AS / 2 S.M.Chappell Lesson 11 2013-2014 際際滷 10 of 12 A cannon fires a steel ball at 120 ms- 1 in at an elevation of 30o from the top of a cliff 80m high. Find the maximum range of the canon. (g = -9.81ms-2) Solving projectile problems 2 120 ms-1 Split the balls initial velocity into horizontal and vertical components 30o 120 ms-1 sin q = 0pp hypsin q x hyp = 0pp sin 30o x 120ms-1 opp Vertical component 0.5 x 120ms-1 = opp Opp = 60 ms-1 vertical component u v Look at the vertical motion in terms of vutas equations.80 m 30o v =u + at Equation 1. s = (v + u ) t Equation 2. 2 s = u t + 遜 at 2 Equation 3. v2 = 2as + u2 Equation 4. Sine rule a = b = c . sin a sin b sin c Cosine rule a2 = b2 + c2 2bc cos A Pythagorus theorem v2 = 2as + u2 v2 = 2 x -9.81 x -80 + (+60ms-1)2 vv = uv = + 60m s-1 t = av = -9.81m s-2 sv = -80 m v2 = 1569.6 +3600 v2 = 1569.6 +3600 v2 = 5169.6 v2 = 5169.6 vv = -71.89 ms-1 -71.89 ms-1 ? ? Minus because it is the final velocity is downwards v =u + at Equation 1 v u = t a -71.89 (+60) = t -9.81-71.89 60 = t -9.81 -131.89 = t -9.81 t = 13.44 s 1.21 s Horizontal component vh = uh = ? t = 13.44 s ah = 0 m s-2 sh = ? cos q = adj . hyp cos 30o = adj . 120 ms-1 cos q x 120 = adj . 0.866 x 120 = 103.9 ms-1 103.9 ms-1 ? s = 103.9 m x 13.44 s s = 1396.4 m 103.9 ms-1 s = v t
- 11. KYS Malaysia Physics AS / 2 S.M.Chappell Lesson 11 2013-2014 際際滷 11 of 12 Problem for you to solve A child throws a stone at a 40o angle from the horizontal. If it reaches the sea 100m below, 70m out, 1.4s later. What was the speed it left his hand at ? (Assume g= 9.81 ms-2) SMCs tip: draw the problem out SMCs tip: write out vutas down the side and assign values vv = ? uv = ? t = 1.4 s av = -9.81 m s-2 sv = ? + -100m vh = uh uh = vh t = 1.4 s ah = 0 m s-2 sh = 70m SMCs tip: Draw out the component vector diagram for the initial vector 40 o v component h component We assume nothing effects the horizontal component we have a displacement and time, v =u + at Equation 1. s = (v + u ) t Equation 2. 2 s = u t + 遜 at 2 Equation 3. v2 = 2as + u2 Equation 4. Sine rule a = b = c . sin a sin b sin c Cosine rule a2 = b2 + c2 2bc cos A Pythagorus theorem s = vh t 70 m-100m -9.81 ms-2 40 ou 1.4 s 70 m = vh = 50ms-1 1.4s 50ms-1 cos 40o = 50ms-1 hyp hyp = 50ms-1 Cos 40o u = 65.3 ms-1 Velocity the stone left the hand at Horizontal component Why is the actual velocity it leaves the hand likely to be slightly higher ? For the stone to reach the distance indicated the velocity will have be higher due to air resistance slowing down the horizontal vector for velocity which we assumed was the same value 50ms-1
- 12. KYS Malaysia Physics AS / 2 S.M.Chappell Lesson 11 2013-2014 際際滷 12 of 12 Problem for you to solve 2 A child throws a stone at a 45o angle from the horizontal. If it reaches the sea 120 m below, 120m sea,170m out, 2.7s later. What was the speed it left his hand at ? (Assume g= -9.81 ms-2) SMCs tip: draw the problem out SMCs tip: write out vutas down the side and assign values vv = ? uv = ? t = 2.7 s av = -9.81 m s-2 sv = ? + -120m vh = uh uh = vh t = 2.7 s ah = 0 m s-2 sh = 170m SMCs tip: Draw out the component vector diagram for the initial vector 45 o v component h component We assume nothing effects the horizontal component we have a displacement and time, v =u + at Equation 1. s = (v + u ) t Equation 2. 2 s = u t + 遜 at 2 Equation 3. v2 = 2as + u2 Equation 4. Sine rule a = b = c . sin a sin b sin c Cosine rule a2 = b2 + c2 2bc cos A Pythagorus theorem s = vh t 170 m-120m -9.81 ms-2 45 ou 2.7 s 170 m = vh = 63 ms-1 2.7s 63 ms-1 63 ms-1 cos 45o = 63ms-1 hyp hyp = 63ms-1 Cos 45o u = 89.1 ms-1 Velocity the stone left the hand at Horizontal component Calculate the height the stone falls from above the cliff. For this vv = 0ms-1 Why? The point where the stones starts its minus velocity due to av uv sin 45 o = opp 89.1 m sin 45 o x 89.1= 63 ms-1 0 2 632 =s 2 x -9.81 v2 = 2as + u2 v2 u2= s 2a s= 202.3m