Revision tips and key areas Higher Physics Revision tips and key areas 24th April You will need paper, and a pen Use the chat pod to ask questions

It’s all in the preparation! Make a study Planner Share it with your parents Trust it! Review each topic Sort LO’s into easy, hard, impossible!!!! Move your phone and other devices into another room Tell your parents what you’ve done Review with notes and scholar Past Papers Make sure you keep the easy, easy If you have time tackle the impossible

Projectiles

Information from the question Initial Velocity of the golf ball 25 ms-1 at 37o above horizontal a) uhoriz = u cos Θ uhoriz = 25 cos 37 uhoriz = 20 m s-1 (1) b) uvert = u sin Θ uvert = 25 sin 37 uvert = 15 m s-1

Part c) Find maximum height of ball. What do we know? Vertically u = 15 ms-1 a = -9.8 ms-2 s = ? What else? v = 0 v2 = u2 + 2 as 02 = 152 + 2 x -9.8 x s s = 11.5 m Height of hilltop = 11.5 - 4 = 7.5 m (4)

Last part

This is about what we understand Any value less than the 7.5 m, calculated in 1.c). Justification; max height will be less since the frictional forces will reduce velocity of ball more than before. Or ball will reach highest point earlier in flight Or time in air will be less Or maximum height reached will be less (2)

Forces Two boxes on a smooth (frictionless) horizontal surface are joined together by a short length of rope, as shown in the diagram below.                                    Box A has mass 2.1 kg and box B has mass 6.3 kg. Box B is being pulled by a horizontal force of 14 N in the direction shown. Calculate the acceleration of the blocks. Calculate the tension in the rope connecting A and B.

Solutions F = 14 N m1 = 2.1 kg m2= 6.3 kg F=ma 14 = 8.4 x a a = 1.7 ms-2 a = 1.7 m = 2.1 F = ma F = 1.7 x 2.1 F = 3.5 N

Free body diagram A skier, mass 80 kg, is skiing down a slope that is inclined at 38° to the horizontal. The frictional force acting on the skier is 175 N. Calculate the acceleration of the skier down the slope.

Result Calculate the component of the skier's weight which acts in the direction parallel to the slope, mgsinQ = 482.7 N Fun = down slope + friction =482.7 – 175 = 307.7 Fun = ma 307.7 = 80 x a a = 3.8

Static Equilibrium A child pushes a toy cart with a force F = 5.1 N. The force is applied at an angle of 15° above the horizontal. The mass of the cart is 3.9 kg Calculate the acceleration of the cart. Calculate the kinetic energy of the cart when it is travelling at 3.1 m s-1.

Answers Calculate the acceleration of the cart. Fh = F cosQ Fh = 5.1 cos 15 = 1.3 ms-2 Calculate the kinetic energy of the cart when it is travelling at 3.1 m s-1. Ek = ½ mv2 = 0.5 x 3.9 x 3.12 = 19 J

Any Questions

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