Linear Momentum, Center of Mass, Conservation of Momentum, and Collision.

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1 PHYS1110H, 2011 Fall. Shijie Zhong Linear Momentum, Center of Mass, Conservation of Momentum, and Collision. Linear momentum. For a particle of mass m moving at a velocity v, the linear momentum for the particle p = mv. (1) Newton s 2 nd law can be expressed in terms of rate of change of linear momentum, as discussed in note 2 on Newton s laws. F net = ma = dp /dt. (2) For a system of N particles, one may consider each particle separately, for example, for j-th particle, F j = d p j /dt. (3) The force may be divided into internal and external forces, where internal force results from interaction among the particles, but external forces are from outside the system of particles: int F j + ext F j = d p j /dt. (4) Summing together for all the particles leads to int F j + ext F j = d p j /dt. (5) According to Newton s 3 rd law, the internal forces between particles are in pairs with equal magnitude but opposite signs, so their sum must be zero. If we define total external force and total momentum for the system as ext F ext = F j, we have P = p j = m j F ext = d P /dt. (7) v j, (6) Note 7 1

2 Conservation of Momentum For a single particle, if the net force acting on the particle is zero, then the momentum of the particle is conserved according to (2) (or constant motion). More interesting is the conservation of momentum for a system of particles. If the total external force acting on the system of particles is zero, then the total momentum of all the particles in the system is conserved from (7). P = p j = m j v j = const. (8) Example 1. Spring-gun recoil A loaded spring-gun is placed on a frictionless surface and is initially at rest. Then the gun fires a marble at angle θ from the surface. The velocity of the marble relative to the gun is v 0. What s the velocity of the gun? Suppose that the gun has a mass M and the marble m. Answer: The gun recoils after firing the marble. Since there is no external force acting on the system of gun-marble in the horizontal direction, the total momentum of the gun-marble is conserved in the horizontal direction, i.e., before and after the firing. The total momentum of the gun-marble system is zero before the firing, as they both are at rest. Suppose that recoil velocity of the gun is V f. The horizontal component of the marble relative to the ground is v 0 cosθ- V f. Conservation of momentum for the system: 0 = MV f + m(v 0 cosθ V f ). Therefore, V f = mv 0 cosθ /(M + m). Note 7 2

3 Collisions and Conservation Laws Consider two particles with masses m 1 and m 2 at velocities v 1 and v 2, colliding with each other. If there is no external force acting on them (the force between them is internal), then the total momentum is conserved, according to (8). m 1 v 1 + m 2 v 2 = m 1 v ' 1 +m 2 v ' 2, (9) where the primed velocities are after the collision. If the collision is elastic, then the kinetic energy is also conserved 1 2 m 2 1 1v m 2 1 2v 2 = 2 m 1v 1 ' m 2v 2 ' 2. (10) If the collision is inelastic, then some of kinetic energy will get lost and create heat, Q (>0). 1 2 m 1 v m 2 v 2 2 = 1 2 m 1 v' m 2 v' 2 2 +Q. (11) Example 2. Elastic collision of two balls. m 2 =3m 1, and v 1 = v 2. What are the velocities after collision? Answer: Using conservation laws of energy and momentum (10) and (11), and assuming that v 1 takes the positive direction, m 1 v 1 3m 1 v 1 = m 1 v' 1 +3m 1 v' 2, 2 m 1 v m1 v 1 = 2 2 m1 v' 1 +3m1 v' 2. v' 1 +3v' 2 = 2v 1, v' v'2 2 = 4v1 2. Simplify them. Substituting v1 into the second equation, 2 4v v1 v' 2 +9v' 2 +3v'2 = 2 4v1, or v' 2 (v 1 + v' 2 ) = 0. This leads to two solutions for mass 2: either v' 2 = 0 or v' 2 = v 1, each leading to velocity for mass 1 either v' 1 = 2v 1 or v' 1 = v 1. Notice that the solution v' 1 = v 1 and v' 2 = v 1 is the same as those before the collision which is always the solution, not interesting. The second set solution v' 1 = 2v 1 and v' 2 = 0 is useful. Note 7 3

4 Center of Mass In (6) and (7), we group the system of particles together and consider their external forces and momentum collectively. We may follow this logic, and write down Newton s 2 nd law for the system of particles with all the masses of the particles adding together as M = m j, F ext = MR, (12) where R is yet defined. Combining (7) and (12), MR = dp /dt = d( m j r j )/dt = m j r j, (13) which indicates that R = 1 N M m jr j, (14) j=1 that is, R is a vector that points to the center of the mass for the system. For distributed mass, for example, an irregularly shaped asteroid, R = 1 r dm = 1 r ρdv, (15) M M where the integral is a volume integral and ρ is the density of the element. Example 3. Center of mass for two balls connected by a rod. The two balls have mass m 1 and m 2, and the rod s mass can be ignored. Answer: Position vectors r 1 and r 2 are given in the figure on the left. According to (14), R = m 1r 1 + m 2 r 2, as shown in the figure. It can be showed that R vector lies on the line connecting m 1 and m 2. Note 7 4

5 r ' 2 = r 2 R = r 2 m 1r 1 + m 2 r 2 m = 1 ( r 1 r 2 ). Here, we consider r 2 R = r ' 2, according to vector subtraction rule. This suggests that r ' 2 must be parallel to vector ( r 1 r 2 ), which indicates that vector R lies on the line connecting m 1 and m 2. Similarly, we can show that r ' 1 = r 1 R m = 2 ( r 1 r 2 ). If the distance between the two masses is l, the distances between the center of mass and m1 and m2 are, respectively, r ' 1 = m 2 l, and r ' 2 = m 1 l. From Newton s law for the system of two masses, (10), F ext = m 1 g + m 2 g = ( ) R, R = g. That is, the center of mass s motion is that of a constant acceleration g, although each of the masses can have much more complicated motion. Example 4. Center of mass for a non-uniform rod. A rod of length L has a non-uniform density. The rod s lighter end is placed at x=0, and the other end is at x=l. The mass per unit length, λ, (also called linear density) is λ=λ 0 x/l. Find the center of mass for the rod. Answer: We need to use (15). First find out the total mass of the rod, M. L L x M = λdx = λ0 0 L dx 1 = 0 2 λ 0 L. R = 1 1 r ρdv = M 1/2λ 0 L xλ 0 x / Ldx e ˆ x = 2 3 Lˆ e x. Collisions and Center of Mass Coordinates Again consider two particles with masses m 1 and m 2 at velocities v 1 and v 2 (in the lab reference frame) colliding with each other with no external force, but we will consider the problem in center of mass (CM) reference frame. First, the CM velocity V in lab frame is Note 7 5

6 V = m 1v 1 + m 2 v 2. (16) V lies on the line connecting velocities v 1 and v 2 (see the above figure). The velocities for masses m 1 and m 2 in CM reference frame are v 1c and v 2c, and are given as: v 1c = v 1 V m = 2 ( v 1 v 2 ), (17) v 2c = v 2 V m = 1 ( v 1 v 2 ). (18) m 1 + m 2 v 1c and v 2c lie back to back along the relative velocity vector v = v 1 v 2 (see the above figure). Notice that (16)-(18) can be obtained by doing timederivatives to positive vectors in Example 3 above. The momenta in the CM frame are: p 1c = m 1 v 1c = m 1m 2 ( v 1 v 2 ) = µ v, (19) p 2c = m 2 v 2c = m 1 m 2 ( v 1 v 2 ) = µ v, (20) where µ = m 1 m 2 /( ) is called reduced mass of the system. The total momentum in CM frame which is the sum of (19) and (20) is zero. This is expected. Because there is no external force to the system, from (12), the CM velocity V = R must be constant with time before and after the collision. The total momentum in the lab frame is m 1 v 1 + m 2 v 2 = ( ) V, (21) Note 7 6

7 which is obviously constant due to conservation of momentum. In CM frame, CM velocity itself is zero, so is the total momentum in CM frame. Because V is constant before and after collision, and also because in CM frame, the velocities for the two masses are in the opposite directions. This makes it convenient to describe collision in CM frame. In the lab frame, the velocities before and after collisions can have very different directions (see figure below), but in the CM frame, v 1c and v 2c before the collision lie back to back along the relative velocity vector v = v 1 v 2, and v ' 1c and v ' 2c after the collision lie back to back along the relative velocity vector v '= v ' 1 v ' 2 (see the figure below). If the collision is elastic, in the CM frame, the magnitudes of velocities for m1 and m2 before and after collisions remain the same, i.e., v 1c = v ' 1c and v 2c = v ' 2c (see the right figure above). The angle is called scattering angle. This can be proved by considering the conservations of momentum and energy in the CM frame. 1 2 m 1 v 1c m 2 v 2c 2 = 1 2 m 1 v' 1c m 2 v' 2c 2. (22) m 1 v 1c m 2 v 2c = 0. (23) m 1 v' 1c m 2 v' 2c = 0. (24) From these three equations, we get: v 1c = v' 1c, and v 2c = v' 2c. (25) Impulse Newton s 2 nd law in equation (2), i.e., the differential form of forcemomentum relation, can also be written as Note 7 7

8 t F dt = P (t) P (0). (26) 0 t F dt is called impulse and is the change of momentum from over time t. 0 When the time is sufficiently short for the momentum change, for example, in a collision process, the impulse or the integral, to a good approximation, is also equal to F Δt. Example 5. A ball rebounded on the floor. A ball of 0.2 kg falls vertically at a speed of 8 m/s immediately before hitting the floor, and is bounced back at the same speed. The ball is in contact with the floor for a duration of 10-3 second. What s the average force that the floor acts on the ball? Answer: Use (26), F Δt = mv a mv d, where the two velocities are ascending and descending velocities, respectively. Take a coordinate system such that the positive is upward, and both the force and ascending velocity are positive. F = 2mv a /Δt = /10 3 = 3200 N. We ignored the gravitational force of the ball when it hit the floor, but it is very small (mg). Momentum and the flow of the mass rocket-propelling problem A rocket can lift off from the ground and send spacecraft to the space because the rocket burns fuel and produces gas that gets expelled from the tail of rocket, thus pushing the rocket upwards/forward. It is essential to have the gas expelled from the rocket s tail, as one can see from the following argument and diagram. The gas is produced in a chamber in the rocket from burning fuel. If the chamber is closed, the generation of the gas increases the pressure in the chamber, producing forces in all directions against the walls. However, the net force on the chamber and hence the rocket is ZERO, as the forces cancel out. However, if one side of the chamber is open, the gas produces forces against the closed sides, resulting in net force pushing the chamber/rocket to Note 7 8

9 one direction. The bottom row shows the forces acting on the gas from the chamber, according to Newton s 3 rd law. This shows that for the open chamber, there is a net force acting on the gas from the chamber to expel the gas out of the chamber. Figure for rocket at time t and time t+δt. Figure for forces on gas and chamber Consider the right figure. For a rocket at time t with mass M and fuel Δm both of which move at a velocity v. Between times t and t+δt, the fuel Δm is burned and expelled at a velocity u relative to the rocket, and the rocket s velocity is boasted to v+δv. The initial and final momentum for the rocketfuel system are P (t) = (M + Δm) v, (27) P (t + Δt) = M ( v + Δ v ) + Δm( v + Δ v + u ). (28) The change in momentum, after dropping out second order term, ΔmΔ v, is Δ P = P (t + Δt) P (t) = MΔ v + Δm u. (29) Note 7 9

10 dp dt = lim Δ P Δt = M d v dt + u dm dt. (30) Notice that for external force F, F = d P dt, dm dt = dm dt, (31) F = M d v dt u dm dt. (32) Notice that u is opposite to v, and u is a constant by design of a rocket. Example 6. A rocket with no external force. Show control on rocket motion. Answer: Use (19) with F=0, and we have dv t dt = u dm M dt, or dv f t dt dt f 1 dm = u dt. M dt v f v i = u ln M f = u ln M i. M i M f t i t i Since M i >M f, u has an opposite direction to v, the right hand side is along the direction of v. Therefore, the rocket motion always increases or v f > v i. Notice that the final velocity is only dependent on u and the ratio of initial to final masses. Example 7. A rocket under gravitational force (i.e., at lift-off). Answer: From (32), Mg = M d v dt u dm dt. dv dt = u dm + t d f t v g, or M dt dt dt f 1 dm = u dt + g (t f t i ). M dt v f v i = u ln M i + g (t f t i ). M f For t i =0, and v i =0, and taking positive direction upward, v f = u ln M i M f gt f. t i Now a lot of the fuel has to be consumed to overcome gravitational force. t i Note 7 10

Chapter 9. Linear Momentum and Collisions

Chapter 9. Linear Momentum and Collisions Chapter 9 Linear Momentum and Collisions Momentum Analysis Models Force and acceleration are related by Newton s second law. When force and acceleration vary by time, the situation can be very complicated.