Formula Sheet for Stage 6 Physics

Formula Name Comments Typical Problem Typical Answerv av=∆ r Average v av = average velocity (m/s) How much distance isrtVelocity r = distance covered (m) covered by a car travellingvav = ∆ ∴r= vav× ∆t= 16 × 4 × 60 = 384mt8.4.1 t = time (s)for 4 mins at an averagera avr∆V=∆tAverageAcceleration8.4.2r r∑ F = ma Newton’sSecond Law8.4.2E k= 1 2 mv 2rp = mvrImpulse = rF tF = G m 1m 2d 2r 3T 2= GM4π 2KineticEnergy8.4.3Momentum8.4.4Impulse8.4.4UniversalGravitation8.5.4Kepler’s ThirdLaw8.5.4a r av= average acceleration (ms -2 )δv = change in velocity (m/s)δt = change in time (s)∑ F r = sum of all forces (N)m = mass (kg)a r = acceleration (ms -2 )E k = kinetic energy (J)m= mass (kg)v = speed (m/s)P = momentum (Ns, kgm/s)m = mass (kg)v = velocity (m/s)Impulse = change in momentum(Ns, kgm/s)F = force (N)G = universal gravitationconstant (6.67 X 10 -11 Nm 2 kg -2 )m 1 = mass of body 1 (kg)m 2 = mass of body 2 (kg)d = separation between the twobodies (m)r = radius of motion (m)T = period of motion (s)G = universal gravitationconstant (6.67 X 10 -11 Nm 2 kg -2 )M = mass of system (kg)speed of 16 m/s?A train accelerates from12 m/s to 18 m/s in 12 s.What is the value of thisacceleration?A ball possesses 10 J ofenergy. What is its mass ifit is moving at 2 m/s?What is the momentum of1400 kg car moving at 6m/s?A ball of mass 0.5 kgtravelling at 3 m/s hitswall and bounces back atthe same speed. If it is incontact with the wall for0.1 s, what is the forceexerted by the wall?What is the gravitationalforce between the Earth(m=6X10 24 kg) and 2 kgball given that the radiusof the Earth is 5.8X10 6 mWhat is the period of a10 12 kg comet that orbits at5 X 10 8 m?ra avr∆V18 −12−= = = 0.5ms∆t121 2E2 × 10E = mv ∴m=22v 2r rp = mv = 1400 × 6 = 8400kgms2kk= =22−15kgImpulse = pf− pi= m( vf− vi) = 0.5(2 − ( −2))r Impulse 2Impulse = Ft ∴ F = = = 20Nt 0.1−1124m1m26.67×10 X 2X6X10F = G == 19. 6N26 2d (5.8×10 m)rTGM∴T=24π24πrGM4π6.67X10X 5X10332= =2−118X1012= 17203s