Presentation on theme: "Taking vertically up as the positive direction:"— Presentation transcript:
1Taking vertically up as the positive direction: SUVAT: A ball is thrown vertically up at a speed of 11.2 ms-1, the hand of the thrower being 1.7 m above the ground when the ball is released. Find the time taken for the ball to hit the ground.Taking vertically up as the positive direction:AB1.7Consider the motion:A Bv =u =a =s =t =11.2–9.8–1.7s u t +a t212=Using:–1.7 = 11.2 t +12( –9.8 ) t24.9 t2 – 11.2 t – 1.7 = 049 t2 – 112 t – 17 = 0177The time taken is( = 2.43 )seconds( 7t – 17 )( 7t + 1 ) = 0
2By Conservation of Momentum: (7 × 3) + (1 × –1) = 3w + 2w 21 – 1 = 5w Momentum : Two particles A and B of masses 3kg and 1kg respectively are moving towards each other along the same straight line, with speeds 7ms–1 and 1ms–1 respectively. After impact the particles move in the same direction with the speed of B being twice that of A. Find the magnitude of the impulse given to A by B.31BABefore:After:71w2wBy Conservation of Momentum:(7 × 3)+ (1 × –1) =3w+ 2wNote the positive direction is taken as the direction that particle A has received an impulse (we want the impulse on A) i.e. to the left.21 – 1 = 5ww = 4Impulse on A =Change in momentum= “mv – mu”I =3(4) –3(7)= –9= 9i.e. Mag of Impulse = 9 Ns
4Pulleys: Particles A and B, of masses 2 kg and 5 kg respectively, are connected by a light inextensible string, passing over a smooth pulley, fixed to the top of a smooth plane, inclined at 30 to the horizontal as shown. Particle A moves on a line of greatest slope on the plane. The system is released from rest with the string taut.Find in terms of g:i) The acceleration of the system.ii) The tension in the string.30°BA
5Now apply Newton’s 2nd Law: 2g 5g 30°BATWe need to completethe force diagram;aa30°Now apply Newton’s 2nd Law:2g5g‘F = ma’, for B:5g – T = 5a …(1)‘F = ma’, for A:T – 2g sin 30 = 2a …(2)Adding (1) + (2):5g – 2g sin 30 = 7a5g – g = 7aSub into (2):
7Moments: A uniform beam AB, of length 6 m and mass 40 kg, is at rest on 2 supports at P and B, where AP = 1 m. A mass of 20 kgis placed on the beam at a distance x m from B, such that the normal reaction at P is double the reaction at B. Find the distance x.2NNAB1P2340g20gxThe beam is in equilibrium, so resolving:3N = 60gN = 20gB :GB =2N × 5– 40g × 3– 20g × xIn equilibrium, GB = 00 = 200g – 120g – 20g xx = 4Note: The position of the mass was not known, and it turns out to be to the left ofthe centre of the rod.