* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Momentum and Impulse
Elementary particle wikipedia , lookup
Monte Carlo methods for electron transport wikipedia , lookup
Brownian motion wikipedia , lookup
Tensor operator wikipedia , lookup
Faster-than-light wikipedia , lookup
Symmetry in quantum mechanics wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Velocity-addition formula wikipedia , lookup
Laplace–Runge–Lenz vector wikipedia , lookup
Quantum vacuum thruster wikipedia , lookup
Uncertainty principle wikipedia , lookup
Relativistic quantum mechanics wikipedia , lookup
Seismometer wikipedia , lookup
Atomic theory wikipedia , lookup
Mass versus weight wikipedia , lookup
Mass in special relativity wikipedia , lookup
Photon polarization wikipedia , lookup
Classical mechanics wikipedia , lookup
Accretion disk wikipedia , lookup
Electromagnetic mass wikipedia , lookup
Centripetal force wikipedia , lookup
Equations of motion wikipedia , lookup
Angular momentum wikipedia , lookup
Center of mass wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Angular momentum operator wikipedia , lookup
Work (physics) wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Matter wave wikipedia , lookup
Classical central-force problem wikipedia , lookup
Specific impulse wikipedia , lookup
Relativistic mechanics wikipedia , lookup
If you apply a force to an object, the effect it has depends on the mass of the object and for how long you exert the force. You apply a constant force, F Newtons for a time t seconds in the direction of motion to a particle of mass m kg. The particle accelerates at a ms-2. The velocity of the particle increases from u ms-1 to v ms-1. You have v = u + at 𝐚= 𝒗−𝒖 𝑡 From Newton’s second law you know that F = ma 𝑚(𝒗 − 𝐮) 𝐅= 𝑡 Ft = mv – mu Because force is a vector quantity, impulse is also a vector quantity The IMPULSE of a force F applied for a time t is defined as Ft Impulse = force x time = Ft The SI unit of impulse is the newton second (Ns) For a particle of mass m moving with velocity v, Momentum = mv The SI unit of momentum is the newton second (Ns) In general Impulse = change in momentum Ft = mv – mu Example 1 A spacecraft of mass 120kg is travelling in a straight line at a speed of 4ms-1. Its rocket is fired for 5 seconds, exerting a force of 150N. Find the new velocity of the spacecraft if the force was directed: (a) in the direction of motion (b) in the direction opposite to motion (b) (a) 150N u = 4ms-1 150N u = 4ms-1 Impulse = Change in Momentum Ft = mv – mu Impulse = Change in Momentum Ft = mv – mu 150 x 5 = 120v – 120 x 4 v = 10.25ms-1 -150 x 5 = 120v – 120 x 4 v = -2.25ms-1 Example 2 A batsman hits a ball of mass 0.15kg at a speed of 40ms-1. The ball moves in a straight line with constant speed until it strikes a net at right angles and is brought to rest in 0.5 seconds. Find the magnitude of the impulse and the average force exerted by the netting on the ball. 40ms-1 F 0ms-1 Mass = 0.15kg Momentum = mass x velocity Momentum = mass x velocity Momentum = 0.15 x 40 Momentum = 0.15 x 0 Momentum = 6Ns Momentum = 0Ns Impulse = Change in Momentum = 6Ns Impulse = Force x time 6 = F x 0.5 F = 12N Example 3 A particle of mass 2kg, travelling with a velocity of (3i + 5j)ms-1, is given an impulse of (2i – 4j)Ns. Find its new velocity. Impulse = Change in Momentum Impulse = mv – mu 2i – 4j = 2v – 2(3i + 5j) 2i – 4j = 2v – 6i – 10j 8i + 6j = 2v v = (4i + 3j)ms-1 Example 4 Steve kicks a ball of mass 0.8kg along the ground at a velocity of 5ms-1 towards Monica. She kicks it back towards him, but lofts it so that it leaves her foot with a speed of 8ms-1 and with an elevation of 40° to the horizontal. Find the magnitude and direction of the impulse of Monica’s kick. (-8cos40i + 8sin40j) ms-1 8ms-1 5i ms-1 5ms-1 Mass = 0.8kg 40° Mass = 0.8kg Impulse = Change in Momentum Impulse = mv – mu Impulse = 0.8(-8cos40i + 8sin40j) – (0.8 x 5i) Impulse = -4.9027i + 4.1138j – 4i Impulse = -8.9027i + 4.1138j Example 4 Steve kicks a ball of mass 0.8kg along the ground at a velocity of 5ms-1 towards Monica. She kicks it back towards him, but lofts it so that it leaves her foot with a speed of 8ms-1 and with an elevation of 40° to the horizontal. Find the magnitude and direction of the impulse of Monica’s kick. Impulse = -8.9027i + 4.1138j I 4.1138 θ 8.9027 I= 8.90272 + 4.11382 𝐈 = 𝟗. 𝟖𝟏𝐍𝐬 4.1138 Tan𝜃 = 8.9027 𝜽 = 𝟐𝟒. 𝟖° Conservation of linear momentum We will now study the motion of systems comprising of two or more bodies, which are free to move separately A system consists of two bodies, A and B, which have momentum MA and MB respectively The total momentum of the system = (MA + MB) Ns Body A now exerts a force, F N on body B for a time t seconds. Body B receives an impulse J = Ft Ns By Newton’s third law, B exerts a force, -F N on A, also for time t secs Body A receives an impulse –J Ns. We know that Impulse = change in momentum The new momentum of A = (MA – J) Ns The new momentum of B = (MB + J) Ns The total momentum of the system = (MA – J) + (MB + J) = MA + MB i.e. the total momentum has not been changed The forces described here are internal forces to the system. The total momentum of a system can only be changed by the application of an external force. This is the principle of conservation of linear momentum. The principle of conservation of linear momentum The total momentum of a system in a particular direction remains constant unless an external force is applied in that direction In the system below: Total momentum before collision = Total momentum after collision mAuA + mBuB = mAvA + mBvB Example 5 A particle of mass 4kg, travelling at a speed of 6ms-1, collides with a second particle, of mass 3kg and travelling in the opposite direction at a speed of 2ms-1. After the collision, the first particle continues in the same direction but with its speed reduced to 1ms-1. Find the velocity of the second particle after the collision. Before A 6ms-1 4kg After A 2ms-1 B 3kg 1ms-1 mAuA + mBuB = mAvA + mBvB (4 x 6) + (3 x -2) = (4 x 1) + 3v 18 = 4 + 3v v = 14/3 ms-1 B v Example 6 A body of mass 5kg, travelling with a speed of 6ms-1, collides with a body B, of mass 3kg, travelling in the same direction with a speed of 4ms-1. On colliding, the two bodies coalesce. Find the velocity of the combined body after the collision. Before A 6ms-1 B 5kg After 3kg A+B v 8kg mAuA + mBuB = mA+Bv (5 x 6) + (3 x 4) = 8v 42 = 8v v = 5.25ms-1 4ms-1 Example 7 A railway truck of mass 4 tonnes, travelling along a straight horizontal trail at a speed of 4ms-1, meets another truck, of mass 2 tonnes, travelling in the opposite direction at a speed of 5ms-1. They collide and become coupled together. Find their velocity after the collision. Before A 4ms-1 5ms-1 4000kg After B 2000kg A+B v 6000kg mAuA + mBuB = mA+Bv (4000 x 4) + (2000 x -5) = 6000v 6000 = 6000v v = 1ms-1 (in direction of A originally) Example 8 A body of mass 4kg travelling with a velocity of (3i + 2j)ms-1 collides and coalesces with a second body of mass 3kg travelling with velocity (i – 3j)ms-1. Find their common velocity after impact. mAuA + mBuB = mA+Bv 4(3i + 2j) + 3(i – 3j) = 7v 15i – j = 7v v= 𝟏𝟓 𝒊 𝟕 𝟏 𝟕 − 𝒋 ms-1 Example 9 Two particles, A and B, of mass 3kg and 2kg respectively, lie at rest on a smooth horizontal table. They are connected by a light inextensible string which is initially slack. B starts to move at a speed of 8ms-1 in the direction AB. Find the common velocity of the particles immediately after the string becomes taut, and the impulse received by each of the particles. Before After A B 3kg 2kg A 8ms-1 B Momentum = 2 x 8 = 16Ns Before v 3kg 2kg Impulse = Change in momentum (momentum after – momentum before) (consider just one of the particles) A = (3 x 3.2) – (3 x 0) = 9.6Ns B = (2 x 3.2) – (2 x 8) = -9.6Ns Impulse = 9.6Ns Momentum = 5v After Momentum Momentum = Before After 16 = 5v v = 3.2ms-1 Example 10 A bullet of mass 50 grams is fired horizontally from a gun of mass 1kg, which is free to move. The bullet is fired with a velocity of 250ms-1. Find the speed with which the gun recoils. AFTER BEFORE 0ms-1 v 250ms-1 m = 0.05kg m = 0.05kg m = 1kg Momentum = 0Ns Before Momentum Momentum = Before After 0 = 12.5 – v v = 12.5ms-1 m = 1kg Momentum = (250 x 0.05) + (1 x –v) After = 12.5 – v Example 11 A gun of mass 800kg, fires a shell of mass 4kg horizontally at a speed of 400ms-1. The gun rests on a rough horizontal surface with coefficient of friction 0.6. The gun is stationary before the shot is fired. Find the distance the gun will move as a result of firing the shell. AFTER 400ms-1 R BEFORE v u = 0ms-1 0.6R 4g 800g 804g Momentum = (400 x 4) + (800 x –v) After = 1600 – 800v Momentum Momentum = Before After Momentum = 0Ns Before 0 = 1600 – 800v v = 2ms-1 Example 11 A gun of mass 800kg, fires a shell of mass 4kg horizontally at a speed of 400ms-1. The gun rests on a rough horizontal surface with coefficient of friction 0.6. The gun is stationary before the shot is fired. Find the distance the gun will move as a result of firing the shell. R a=? u = 2ms-1 v = 0ms-1 0.6R s 800g F = ma -0.6R = 800a -0.6(800g) = 800a a = -5.88ms-2 Resolving vertically R = 800g u = 2ms-1 v = 0ms-1 a = -5.88ms-2 s = ? v2 = u2 + 2as 02 = 22 + (2 x -5.88 x s) s = 0.34m (34cm)