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EdExcel Mechanics 2 Kinematics of a particle Chapter assessment Take g = 9.8 ms-2 unless otherwise instructed. 1. A golf ball is hit over horizontal ground from a point O on the ground. The velocity of projection is 30 ms-1 at 40° to the horizontal. The effects of air resistance should be neglected. (i) The ball is y m above the ground t seconds after projection. Write down an expression for y in terms of t and hence determine the time at which the ball first hits the ground. [5] The ball passes directly over a tree which is at a horizontal distance of 34 m from O. (ii) Determine the speed of the ball as it passes over the tree. Calculate also the angle between the direction of motion of the ball and the horizontal at that time, making it clear whether the ball is rising or falling. [10] 2. In this question you should take g = 10 ms-2. The effects of air resistance should be neglected. A small stone is fired from a catapult 1 m above horizontal ground at a speed of 30 ms-1. The angle of projection with the horizontal is , where cos 0.6 and sin 0.8 . The stone hits a vertical wall that is a horizontal distance of 27 m from the point of projection. This information is shown in the diagram below together with x- and y-axes and the origin O on the ground; the units of the axes are metres. y wall 30 ms Not to scale -1 1m O 27 m x (i) Show that, after t seconds, the horizontal displacement of the stone from O, x m, and the vertical displacement, y m, are given by x 18t and y 1 24t 5t 2 . (ii) What is the value of t when the stone hits the wall? How high is the stone above the ground when it hits the wall? © MEI, 01/03/10 [5] [3] 1/7 EdExcel M2 Kinematics Assessment solutions (iii)Show that the stone is rising when it hits the wall. [2] (iv) Find the horizontal displacement of the stone when it is at a height of 17 m above the ground. [5] 3. A particle moves on the x-axis. Its displacement, x m, from the origin O is given by x 3t 2 3t 2 , where t is the time in seconds. How far is the particle from O when it is instantaneously at rest? [5] 4. A racing car starts off down a straight section of track towards the first corner. Its speed, v ms-1, is modelled for the first four seconds of its motion by v t 3 9t 2 24t , 0t 4. (i) Find an expression for the distance travelled by the car in the first t seconds. Calculate the distance travelled from t = 2 to t = 4. [5] (ii) Show that the acceleration, a ms-2, of the car at time t is given by a k (t 2)(t 4) , where k is a constant to be determined. [2] 5. The position vector, r, of a particle at time t is given by r t 2i (5t 2t 2 ) j , where i and j are the standard unit vectors, lengths are in metres and time is in seconds. (i) Find an expression for the acceleration of the particle. [4] (ii) Is the particle ever at rest? [2] 6. The velocity, v, of a particle is given as v 2t 2 3t 13 t 3 . (i) Show that the acceleration of the particle is zero when t = 1 and when t = 3. [3] (ii) Calculate the displacement of the particle from its position when t = 1 to its position when t = 2. [4] 7. An insect moves in a straight line. The time, t, is in seconds and distance travelled is in metres. © MEI, 01/03/10 2/7 EdExcel M2 Kinematics Assessment solutions The velocity, v ms-1, of the insect is given by v t 2 4t , 0 t 6, v c, 6 t 10, v at b, 10 t 15. You are also given that v = 4 when t = 12. (i) Show that c = 12. [2] (ii) Calculate the values of a and b and briefly describe the motion of the insect in the interval 10 t 15 . [4] (iii) Calculate the values of v for t = 0, t = 2 and t = 4. Sketch the v-t curve for the motion of the insect in the interval 0 t 6 . [3] (iv) Calculate the distance travelled by the insect in the interval 0 t 6 . [6] Total 70 marks © MEI, 01/03/10 3/7 EdExcel M2 Kinematics Assessment solutions Kinematics of a particle Solutions to Chapter assessment 1. (i) For vertical motion: u 30 sin 40 sy t t g 9.8 s ut 21 at 2 y 30t sin 40 4.9t 2 When the ball hits the ground, y = 0. 0 30t sin 40 4.9t 2 0 t (30 sin 40 4.9t ) 30 sin 40 3.94 (3 s.f.) 4.9 t = 0 is the time of projection, so the time the ball first hits the ground is 3.94 seconds (3 s.f.) t 0 or t (ii) Horizontal speed 30cos 40 22.9813 When it passes over the tree, 34 30t cos 40 t Vertically: u 30 sin 40 34 30cos 40 v u at v v 30 sin 40 34 30 cos 40 a 9.8 t 4.7849 9.8 34 30 cos 40 4.7849 22.9813 Speed 22.9813 2 4.78492 23.5 (3 s.f.) The speed of the ball as it passes over the tree is 23.5 ms-1 (3 s.f.) 4.7849 22.9813 11.8 (to nearest degree) tan The direction of motion of the ball is 11.8° above the horizontal. © MEI, 01/03/10 4/7 EdExcel M2 Kinematics Assessment solutions The ball is rising, since its vertical velocity is positive. 2. (i) Horizontally, speed 30cos 30 0.6 18 Horizontal speed is constant, so x 18t . Vertically, initial speed 30sin 30 0.8 24 Initial height = 1 u 24 s ut 21 at 2 a 10 y 1 24t 5 t 2 s y1 y 1 24t 5 t 2 t t (ii) When the stone hits the wall, x = 27 x 18t 27 18t t 1.5 When t = 1.5, y 1 24 1.5 5 1.5 2 1 36 11.25 25.75 (iii) When it hits the wall, vertically: u 24 v u at t 1.5 a 10 v ? 24 10 1.5 24 15 9 Since the vertical speed is positive when it hits the wall, the stone is rising. (iv) When y = 17, 17 1 24t 5 t 2 5 t 2 24t 16 0 (5 t 4)(t 4) 0 t 0.8 or t 4 Since the stone hits the wall at t = 1.5, the value of t when it is at a height of 17 m is 0.8. When t = 0.8, x 18t 18 0.8 14.4 The horizontal displacement of the stone is 14.4 m. 3. x 3t 2 3t 2 dx 6t 3 dt When particle is instantaneously at rest, 6t 3 0 t 0.5 When t = 0.5, x 3 0.5 2 3 0.5 2 1.25 v © MEI, 01/03/10 5/7 EdExcel M2 Kinematics Assessment solutions It is 1.25 m from O when it is instantaneously at rest. 4. (i) v t 3 9t 2 24t s v dt 41 t 4 3t 3 12t 2 c When t = 0, s = 0 c = 0 s 41 t 4 3t 3 12t 2 Distance travelled from t = 2 to t = 4 is 41 44 3 43 12 42 41 2 4 3 2 3 12 2 2 64 192 192 4 24 48 36 Distance travelled = 36 m. dv 3t 2 18t 24 3 t 2 6t 8 3(t 2)(t 4) dt so k = 3. (ii) a 5. (i) r t 2 i (5 t 2t 2 ) j dr 2t i (5 4t ) j dt dv a 2i 4 j dt v (ii) The velocity of the particle is given by 2t i (5 4t ) j . For the particle to be at rest, both components of the velocity must be zero. The component of the velocity in the i direction is only zero when t = 0, and the component of the velocity in the j direction is only zero when t = 1.25. So the particle is never at rest. 6. (i) v 2t 2 3t 31 t 3 dv 4t 3 t 2 dt When a = 0, t 2 4t 3 0 a (t 1)(t 3) 0 t 1 or t 3 © MEI, 01/03/10 6/7 EdExcel M2 Kinematics Assessment solutions 2 (ii) Displacement v dt 1 2 23 t 3 23 t 2 121 t 4 1 163 6 43 23 23 1 12 13 12 7. v t 2 4t , v c, 0 t 6, 6 t 10, 10 t 15. v at b , (i) When t = 6, v 62 4 6 12 Therefore c = 12. (ii) When t = 10, v = 12 10a b 12 When t = 12, v = 4 12a b 4 Subtracting: 2a 8 a 4, b 52 The insect is decelerating at a constant rate. (iii)When t = 0, v 02 4 0 0 When t = 2, v 2 2 4 2 4 When t = 4, v 42 4 4 0 v 12 8 4 0 2 4 6 t -4 4 4 6 6 (iv) Displacement for 0 t 4 v dt 31 t 3 2t 2 0 0 32 64 3 32 3 Displacement for 4 t 6 v dt 31 t 3 2t 2 4 4 72 72 64 3 32 32 32 1 Total distance travelled = 3 3 21 3 m. © MEI, 01/03/10 32 3 7/7