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Momentum Physicists have defined a quantity called momentum (動量) of a moving object as... Momentum = mass velocity P = mv Unit: kg m s1 A vector quantity 1 Newton’s 3rd law and Conservation of momentum A FAB B FBA Newton’s 3rd law FBA = -FAB m A v A m Au A mB v B mB u B t t mA v A mAu A mB vB mB u B m A u A mB u B m A v A m B v B If there is no external force acting on a system, then the total momentum of the system is conserved 2 Example 1 A bullet of mass 10 g traveling horizontally at a speed of 100 ms-1 embeds itself in a block of wood of mass 990 g suspended by strings so that it can swing freely. Find (a) the vertical height through which the block rises, and (b) how much of the bullet’s energy becomes internal energy. Solution: (a) Let v be the velocity of the block just after impact. By conservation of momentum, (0.01)(100) = (0.01 + 0.99)v 100 ms-1 v = 1 ms-1 By conservation of energy, ½ mv2 = mgh ½ (1)2 = 10h h = 0.05 m h v 3 Example 1 A bullet of mass 10 g traveling horizontally at a speed of 100 ms-1 embeds itself in a block of wood of mass 990 g suspended by strings so that it can swing freely. Find (a) the vertical height through which the block rises, and (b) how much of the bullet’s energy becomes internal energy. Solution: (b) Some of the K.E. is converted into internal energy. Energy required 100 ms-1 = ½ mu2 – ½ (m + M)v2 = ½ (0.01)(100)2 – ½ (0.01 + 0.99)(1)2 = 49.5 J h v 4 Application 1 – measuring the inertial mass Weighing machine or beam balance only measures the weight. To find the mass (gravitational mass), we must depend on the equation m = W/g. However, g varies from place to place and we cannot use this method to find the mass in the outer space. Another way to determine the mass (inertia mass) without depending on the value of g is to use the principle of conservation of momentum. 5 Application 1 – measuring the inertial mass Consider two trolleys of m1 and m2 are in contact. A spring is used to cause them to explode, moving off the velocity v1 and v2. m1 m2 Before explosion v1 m1 m2 v2 After explosion If no external force acts on them, conservation of m2 v1 momentum gives 0 = m1v1 + m2v2 m1 v2 Therefore, by measuring their velocities, the ratio of their masses can be found. If a standard trolley is used, the other mass can be found. 6 Example 2 Compare the masses of 2 objects X and Y which are all initially at rest. They explode apart and their speeds become 0.16ms-1 and 0.96ms-1 respectively. m1 m2 Before explosion 0.16 ms-1 m1 m2 0.96 ms-1 After explosion By conservation of momentum, m2 0.16 1 0 = m1(-0.16) + m2 (0.96) m1 0.96 6 The ratio of the masses m1: m2 = 6:1 7 Applications 2 – Rocket engine The rocket engine pushes out large masses of hot gas. The hot gas is produced by mixing the fuel (liquid hydrogen) with liquid oxygen in the combustion chamber and burning the mixture fiercely. The thrust arises from the large increase in momentum of the exhaust gases. 8 Applications 2 – Jet engine A jet engine uses the surrounding air for its oxygen supply. The compressor draws in air at the front, compresses it, fuel is injected and the mixture burns to produce hot exhaust gases which escape at high speed from the rear of the engine. These cause forward propulsion. 9 Recoil of rifles Two factors affects the recoil speed of a rifle. 1 momentum given to the bullet 2 momentum given to the gases produced by the explosion. gas bullet rifle 10 Example 3 A rife fires a bullet with velocity 900 ms-1 and the mass of the bullet is 0.012 kg. The mass of the rifle is 4 kg and the momentum of gas ejected is about 4 kg ms-1. Find the velocity of the recoil of the rifle. -1 gas (4 kg ms-1) 900 ms v rifle bullet Solution: By conservation of momentum 0 = 0.012 x 900 + (-4) + 4v v = -1.7 ms-1 The recoil velocity is 1.7 ms-1 backward 11 mv mu Rearrange terms in F = t Ft = mv mu impulse: product of force & time during which the force acts (vector) Impulse = change in momentum 12 a force / N Force-time graph of impact force / N time / s time / s Area under F-t graph = impulse = change in momentum 13 Collisions General properties A large force acts on each colliding particles for a very relatively short time. The total momentum is conserved during a collision if there is no external force. In practice, if the time of collision is small enough, we can ignore the external force and assume momentum conservation. For example, when a racket strikes a tennis ball, the effect due to gravitational force (external force) is neglected since the time of impact is very short. 14 Collisions For another example, when a cannon fires a metal ball, the effect due to frictional force (external force) on the cannon is neglected since the time of explosion is very short. 15 Different kinds of collisions Assume no external force acts on colliding bodies. Collisions Momentum conserved? K.E. conserved? Elastic collisions Yes Yes Inelastic collisions Yes No Completely inelastic collision Yes No (K.E. loss is maximum) 16 Relative velocity rule (For elastic collisions only) relative speed before collision = relative speed after collision u1 u2 v1 Before collision v2 After collision u1 – u2 = v2 – v1 17 Find the velocities of A and B after the elastic collision. 5 1 kg ms-1 3 ms-1 v1 2 kg Before collision v2 After collision By conservation of momentum, (1)(5) + (2)(3) = (1)(v1) + (2)(v2) v1 + 2v2 = 11 --(1) By relative velocity rule, 5 – 3 = -(v1 – v2) v2 – v1 = 2 --(2) ∴ v1 = 2.33 ms-1 and v2 = 4.33 ms-1 18 Trolley A of mass m with velocity u collides elastically with trolley B of mass M at rest. Find their velocities v and V after collision. v V At rest u A B A B By conservation of momentum, mu = mv + MV --- (1) By relative velocity rule, u = V – v --- (2) (1) + (2) x m: 2mu 2mu = (m + M)V ⇒ V mM (1) – (2) x M: mM u mu – Mu = (m + M)v ⇒ v mM 19 Trolley A of mass m with velocity u collides elastically with trolley B of mass M at rest. Find their velocities v and V after collision. 2mu mM v mM u V mM At rest u A B A B Some interesting results stop after the collision If m = M, trolley A will ___________________________ If m < M, trolley A will move in the opposite direction after the collision ________________________ remain at rest after the collision If m << M, trolley B will __________________________ 20 Collisions in two dimensions Momentum can be resolved along different directions. Consider the following oblique impact. A particle of mass m with velocity u collides with another stationary particle of mass M obliquely as shown below. y Apply conservation of Mv2 M momentum along x-axis, mu = mv1cos b + Mv2cos a m a mu x Apply conservation of b momentum along y-axis, 0 = Mv2sin a + (-mv1sin b) mv1 21 A stationary object of mass 500 g explodes into three fragments as shown below. Find the speed the two larger fragments v1 and v2 after the explosion. 0.2v2 200 g 200 g 30o 0.2v1 500 45o After explosion g 100 g Before explosion -1 0.1 x 40 ms Along vertical direction: 0 = 0.2 v2 sin 30o – 0.1(40) sin 45o v2 = 28.3 ms-1 Along horizontal direction: 0 = 0.2v2cos30o + 0.1(40)cos45o – 0.2v1 v1 = 38.6 ms-1 22 Right-angled fork track If one particle collides with another identical particle obliquely, the angle between the directions of motion of the two is always 90o if the collision is elastic. Significance Alpha particles travel in a cloud chamber and have collisions with helium atoms. The figure above shows after collision the alpha particle and the helium atom travel at right angle to each other. This implies alpha particles must have the same mass as helium atom and actually they are nuclei of helium. 23 Mathematical proof of right-angled fork track (i.e. q = 90o) u a v2 a collision between identical particles q v1 Along the line of centres, mucosa mv1 cosq mv2 u cosa v1 cosq v2 (1) Perpendicular to the line of centres, musin a mv1 sin q u sin a v1 sin q (2) For elastic collision, 1 1 1 2 2 2 2 2 2 mu mv1 mv 2 u v1 v 2 (3) 2 2 2 24 (1)2 + (2)2: u cos a u sin a v1 cosq v2 v1 sin 2 q 2 2 2 2 2 2 u 2 v1 cos2 q 2v1v2 cosq v2 v1 sin 2 q 2 2 2 u 2 v1 2v1v2 cosq v2 (4) 2 2 Sub.(4) into (3): v1 2v1v2 cosq v2 v1 v2 2 2 2 2 2v1v2 cosq 0 cosq 0 q 90 The two particles must move at right angle to each other. 25 Elastic collision between a smooth ball and a fixed surface Momentum along the fixed surface is unaltered by the impact. v v Force of impact Momentum along the fixed surface is unaltered by the impact. 26 Elastic collision between a smooth ball and a fixed surface Since the collision is elastic, after the impact, it rebounds with the same speed. v q f v Momentum along the fixed surface is unaltered by the impact. mv sin q = mv sin f ⇒q = f ⇒ the angle of reflection = the angle of incident Change in momentum perpendicular to the surface = mv cos q – (-mv cos f) = 2mv cos q. Force of impact = 2mv cos q / t 27