Exam 1 – Review Session

1. Exam 1 – Review Session Clarification: When I write w.r.t , it means “with respect to” PHY231

2. Overview • Wednesday September 30th 9:10-10 am BPS 1410 • Exam 1 includes • Introduction • Motion in 1D • Motion in 2D • Laws of Motion • Multiple choice questions only PHY231

3. Overview • Introduction • Kinematics • Motion of an object under constant acceleration • 1-D and 2-D motion • Laws of motion • Newton’s concept of forces, F=ma • Apply kinematics to determine object’s motion PHY231

4. SI - Units • SI - base units Length metre m Mass kilogram kg Time second s Electric current ampere A Thermodynamic temp. degree Kelvin °K Luminous intensity candela cd Amount of substance mole mol • All other units are derived from these PHY231

5. Trigonometry PHY231

6. Vector components A vector can also be parameterized using its magnitude (I.e. length) and its direction (I.e. angle w.r.t. the coordinate system) Consider The magnitude of V and angle w.r.t. the x-axis are One also has PHY231

7. Quadratic equation • Reminder for equations of the type • (Real) solutions are given by PHY231

8. 2 Equations – 2 unknowns Isolate 1 unknown and plug into 2nd equation 2 equations – 2 unknowns Intermediate calculation Done, verify that it works in initial system of equations !! PHY231

9. Kinematics – constant acceleration • Time interval Change in position Change in velocity • Average velocity Average acceleration • Important kinematic relations PHY231 9

10. 2-D - Projectile motion The initial conditions can be broken down into its x- and y-components The equations of motion become PHY231

11. Projectile motionAt what Dx is the shell hitting hill2? Vi =19 m/s 48º 1 h1=20m 2 h2=13m y Vi =19 m/s vyi =vi*sin48º x 48º vxi =vi*cos48º

13. Area under v(t) is Dx The area under the graph of v(t) is the displacement Dx !! v(m/s) 2 1 0 t(s) 0 2 4 5 • Example • An object has vi=1 m/s • Accel. with a=0.5 m/s2 for 2s • Accel. with a=0.0 m/s2 for 2s • Decel. with a=-2.0 m/s2 for 1s • What is the total displacement of the objects? • Answer : Dx = 8m 1m 4m 2m 1m PHY231

14. The area under the graph of a(t) is the velocity change Dv !! Area under a(t) is Dv • Example • Accel. with a=1 m/s2 for 2s • Accel. with a=0 m/s2 for 2s • Decel. with a=-2 m/s2 for 1s • What is the total change of velocity? • Answer : Dv = 0 m/s a(m/s2) 1 2m/s 5 4 0 0 2 t(s) -2m/s -2 PHY231

15. yG Relative motion xG • A boat is moving on a river • Boat velocity w.r.t. Water • vBW • Water velocity w.r.t. Ground • vWG • Boat velocity w.r.t. Ground • vBG = vBW + vWG • Similar relations exist for displacement instead of velocity vBW vWG vBG PHY231 15

16. yG xG PHY231

17. Newton’s Laws • First Law : If F = 0 then a = 0 and v =constant • Second Law : Fnet = ma • Third Law : FAB = - FBA • First Law: To change the velocity of an object, you must apply a force on it • Second Law: Describes the relation between the force and the acceleration • Third Law: Action-reaction, A applies force F on B necessarily means B applies –F on A PHY231

18. Mass • A measure of the resistance of an object to changes in its motion • The larger the mass, the less it accelerates under the action of a given force • SI unit: kg • Scalar quantity PHY231

19. Tension in a rope • Ignore any frictional effects of the rope • Ignore the mass of the rope • The magnitude of the force exerted along the rope is called the tension • The tension is the same at all points in the rope (magnitude of the tension vector T) • The Tension follows the rope. Draw it at the junction of the object and the rope, pointing AWAY form the object PHY231

20. Sliding box on an inclined frictionless table • What is the box overall acceleration if the angle is equal to 30º ? • Forces acting on the box • Gravity Fg • Normal force n PHY231 20

21. Cart + fan • A cart with a fan mounting on it (1.5 kg total) is on a frictionless inclined surface. The fan can produce a force of Ffan= 5.0 N. What is the angle the inclined surface should have with respect to the horizontal direction so that no net force is acting on the cart (i.e. fan cancels out gravity). g=9.8 m/s2. • A) 10º • B) 20º • C) 30º • D) 45º y x Ffan PHY231

22. y x Ffan -Fgsinq PHY231

23. Friction forces • Force parallel to the surfaces in contact • Static friction • No motion between Surfaces in contact • Max static friction force • n is the normal (perpendicular) force between the surfaces in contact. It is the reaction force of the surface being pushed on • Kinetic friction • Motion between Surfaces in contact • Friction force is constant

24. m Vi=80.8 km/h truck M fstatic • Max friction force • Newton’s law • Under constant acc, we can write -Fbrakes m Vi=80.8 km/h -fstatic M PHY231

25. friction • A mass m1=10 kg is on a table and pulled by a mass m2=20 kg through a rope and a pulley. • m1 is sliding to the right and the coefficient of kinetic friction between m1 and the table is 0.20. • What is the magnitude of the acceleration of either mass? g=9.81 m/s2. y • A) 2.3 m/s2 • B) 5.9 m/s2 • C) 7.3 m/s2 • D) 9.8 m/s2 x PHY231

28. A) 2.3 m/s2 • B) 5.9 m/s2 • C) 7.3 m/s2 • D) 9.8 m/s2

29. Gravitational force • Attractive force between every particles in the universe • Proportional to the product of the masses • Inversely proportional to the square of the distance

30. Gravity on earth At the surface of the earth • ME ~ 6.0 1024 kg • RE ~ 6.4 106 m PHY231 30