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**Taking 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: A B 1.7 Consider the motion: A B v = u = a = s = t = 11.2 –9.8 –1.7 s u t + a t2 1 2 = Using: –1.7 = 11.2 t + 1 2 ( –9.8 ) t2 4.9 t2 – 11.2 t – 1.7 = 0 49 t2 – 112 t – 17 = 0 17 7 The time taken is ( = 2.43 ) seconds ( 7t – 17 )( 7t + 1 ) = 0

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

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**When two particles collide they exert equal and opposite impulses **

Summary of key points: When two particles collide they exert equal and opposite impulses on each other. Impulse = Change in momentum = Final momentum – Initial momentum = mv – mu 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. This PowerPoint produced by R Collins; © ZigZag Education 2008–2010

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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. 30° B A

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**Now apply Newton’s 2nd Law: 2g 5g**

30° B A T We need to complete the force diagram; a a 30° Now apply Newton’s 2nd Law: 2g 5g ‘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 = 7a 5g – g = 7a Sub into (2):

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**To solve problems involving pulleys we apply Newton’s Second **

Summary of key points: 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. This PowerPoint produced by R Collins; © ZigZag Education

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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. 2N N A B 1 P 2 3 40g 20g x The beam is in equilibrium, so resolving: 3N = 60g N = 20g B : GB = 2N × 5 – 40g × 3 – 20g × x In equilibrium, GB = 0 0 = 200g – 120g – 20g x x = 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.

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**The moment of a force about a point is defined as the product of the **

Summary of key points: 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. The clockwise moment of a force of magnitude P, about point A, is given by: A P d GA = Pd 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. This PowerPoint produced by R.Collins; ©ZigZag Education 2010

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ENGR 215 ~ Dynamics Sections 15.1-15.3. Impulse and Linear Momentum In this section we will integrate the equation of motion with respect to time and.

ENGR 215 ~ Dynamics Sections 15.1-15.3. Impulse and Linear Momentum In this section we will integrate the equation of motion with respect to time and.

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