1. Graphs of Motion Physics – Equations of Uniform Motion 11.1.5

2. 80m g = a = 10ms -2 1) i

3. 80m g = a = 10ms -2

4. Simulation results

5. 1) ii 80m a = 10ms -2 u = 20ms -1

6. 0 5 10 15 20 1 2 3 4 Note that since we have neglected air resistance there are no decelerating forces acting on V H as it is horizontal to gravity. Hence, velocity to constant in this direction until it stops falling after 4 seconds

7. 0 10 20 30 40 1 2 3 4 t=0, v = 0 t=1, v = 10 t=2, v = 20 t=3, v = 30 t=3, v = 40 v We can use the formulae v = u+at. This simplifies to v = at (u=0). As gravity takes hold on the ball its velocity increases. If a = 10ms -2 we can simply say that v = 10t

8. 0 20 40 60 80 1 2 3 4 h / m We can use the formulae s=ut+0.5at 2. This simplifies = s=0.5at 2 But if a = 10ms -2 we can simply say that s = 5t 2 (NB s is actually s taken away from 80) t=0, s = 0 t=1, s = 5 t=2, s = 20 t=3, s = 45 t=4, s = 80

9. 2) i

10. time in seconds (s) Distance fallen (m) (ii) To work out the gradient or velocity you take a tangential line which fits the slope at that point i.e. 0.70s  s/  t = 5m / 0.71s = 7.04ms -1 (iii) We can explain the upwards curve as each second the ball falls b y more each second as the ball is accelerating under the force of gravity. (iv) Grad is velocity of ball. To estimate “g” from this experience we can assume that ball started from rest. Hence we can use our formulae for uniformly accelerated motion; v = u+at (v – u)/t = a (7.04ms -1 – 0 ) / 0.7s = 10.057ms -2 a = 10.1 ms -2

11. 2) ii Newton's 3 Laws of Motion I. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it. II. The relationship between an object's mass m, its acceleration a, and the applied force F is F = ma. III. For every action there is an equal and opposite reaction. Answer The second law can be said to apply here as the falling body is falling under constant acceleration of 9.81Nkg -1 or 9.81ms -2. This means that the force applied by this acceleration is; F = ma = constant of 9.81ms -1. This leads us to the commonly used formulae of W = mg. Hence, the force of weight acting on the body does not change throughout the journey. The third law also comes into play as there is an opposite force of drag which is the reaction to the acceleration which quickly increases as the body falls until it is equal and opposite to W. At this point the object must fall at a constant or terminal speed.

12. 2.5m 3) a 20  5.0ms -1 Resolving horizontally requires; R = 5.0ms-1 adj = Rcos  = 5.0ms -1 x cos  = 4.67ms -1 s/v = t = 2.5m / 4.67ms -1 = 0.54s b) Resolving in a vertical direction can be achieved simply using; tan  =opp/adj or adj x tan  =opp Hence; s = opp = tan(20) x 2.5m s = 0.909m s = 0.91m g = a = 10ms -2