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Published byXavier Gladhill Modified about 1 year ago

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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. v = u = a = s = t = Taking vertically up as the positive direction: 11.2 –9.8 –1.7 s u t + a t2a t2 1212 = Using: –1.7 = 11.2 t + 1212 ( –9.8 ) t 2 4.9 t 2 – 11.2 t – 1.7 = 0 49 t 2 – 112 t – 17 = 0 ( 7t – 17 )( 7t + 1 ) = 0 The time taken is seconds 17 7 ( = 2.43 ) Consider the motion: A B A B 1.71.7

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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. w 7 1 2w2w 3 1 B A Before: After: By Conservation of Momentum:(7 × 3)3w3w 21 – 1 = 5w w = 4 Impulse on A =Change in momentum= “mv – mu” 3(4) –= –9= 9 i.e. Mag of Impulse = 9 Ns I =3(7) + (1 × –1) = + 2w

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Summary of key points: This PowerPoint produced by R Collins; © ZigZag Education 2008–2010 When two particles collide they exert equal and opposite impulses on each other. The law of conservation of momentum states that when there are no external forces acting on a system, the total momentum of the system is constant. For a collision: The total momentum before a collision = the total momentum after. Impulse = Change in momentum = Final momentum – Initial momentum = mv – mu

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30 ° B A Pulleys: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.

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30 ° B A T T 2g 5g N a a 30 ° We need to complete the force diagram; Now apply Newton’s 2 nd Law: ‘F = ma’, for B: 5g – T = 5a …(1) T – 2g sin 30 = 2a …(2) Adding (1) + (2): 5g – 2g sin 30 = 7a 5g – g = 7a Sub into (2): ‘F = ma’, for A:

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Summary of key points: This PowerPoint produced by R Collins; © ZigZag Education 2008-2010 To solve problems involving pulleys we apply Newton’s Second Law F = ma, in the direction of the acceleration for each particle. We always assume that the pulley is smooth. This implies that the tension in the string is equal along its whole length. The string is also usually considered to be inextensible. This implies that the connected particles move with the same acceleration.

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Moments: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 kg is 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. 2 1 A B N 2N2N 20g 40g P x 3 The beam is in equilibrium, so resolving:3N = 60g B : G B = 2N × 5– 40g × 3– 20g × x In equilibrium, G B = 00 = 200g – 120g – 20g xx = 4 Note: The position of the mass was not known, and it turns out to be to the left of the centre of the rod. N = 20g

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Summary of key points: This PowerPoint produced by R.Collins; ©ZigZag Education 2010 The moment of a force about a point is defined as the product of the magnitude of the force and the perpendicular distance of the line of action of the force from the point. A P d G A =PdPd A : If a rod is in equilibrium, the resultant force on the rod will be zero, and the sum of the moments about any point will also be zero. The clockwise moment of a force of magnitude P, about point A, is given by:

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