The references:

;/ The Inelastic Collision A completely inelastic collision occurs when two objects stick together. An example of this would be two colliding balls of putty that stick together or two railroad cars that couple together when they collide. The kinetic energy is not necessarily transformed totally to other forms of energy in an inelastic collision. For example, when a travelling railroad car collides with a stationary one, the coupled cars travel off with some kinetic energy. Even though kinetic energy is not conserved in inelastic collisions, the total energy is conserved, and the total vector momentum is always conserved. To be more specific, a railroad car of mass 10 000 kg travelling at a speed of 24.0 m/s strikes an identical car at rest. The cars lock together. We want to determine their common speed afterward. The total initial momentum is myvy4 + movp = (10 000 kg)(24.0 m/s) + (10 000 kg)( 0 m/s) = 2.40 x 10 kg-mis. After the collision, the total momentum will be the same but it will be shared by both cars. Since the two cars become attached, they will have the same velocity, call it v;. (mq + my)v; = 2.40 x 10 kg'm/s 2.40x10kg - m/s v, = > =120 m/s 2.00x10kg Now let us compare the kinetic energies before and after the collision. The kinetic energy before the collision was 2= KE, = my; i %(10 000kg)(24.0m/s) 2 =2.88 x 108 J After the collision, the total kinetic energy is KE, = %(m. +m,)v: = %(20 000kg)(12.0m/s) 2 =1.44 x 105 J Thus the kinetic energy lost and converted to other forms of energy is 2.88x 108 J-1.44x108J=144x1084 my +m; h=0.650 m The physics of the ballistic pendulum can be divided into two parts. First there is the completely inelastic collision between the bullet and the block. Second there is the resulting motion of the block and bullet as they swing upward. The total momentum of the system (block plus bullet) is conserved during the collision. This is because the suspension wire supports the system's weight, which means that the sum of the external forces acting on the system is nearly zero. Furthermore, as the system swings upward, the principle of conservation of mechanical energy applies, because there aren't any non-conservative forces doing work. The tension force in the wire does no work because it acts perpendicular to the motion. Air resistance is negligible during the swing. The total momentum before the collision is just the momentum of the bullet since the block is not in motion: p; = myv4;. The total momentum after the collision is the sum of the momentum of the block and the bullet after they become attached together: ps = (mq + mo)vs. (mq + ma)ve = mqvy; We can solve this equation for the initial speed of the bullet vy;. m +m =vf m Vi To determine v4;, a value is needed for the speed v; immediately after the collision. This value can be obtained from the maximum height to which the system swings by using the principle of conservation of mechanical energy. The total energy before the bullet and the block of wood collide is all kinetic energy: 1 2 5 (m, +my)v; This energy becomes converted to gravitational potential energy at the top of the swing: (m4 + m)gh;. By setting these two energies equal to each other, we can determine v4. v = ghf - '2(9.80m/sz)(0.650m) =13.57m/s The speed of the bullet can now be found. _ 0.0100kg +2.50kg v, = (3.57m/s) = 896 m/s 0.0100kg The Elastic Collision: One Mass at Rest We defined an elastic collision as one in which the total kinetic energy of the system after the collision is equal to the total kinetic energy before the collision. Mathematically this can be written as Since momentum is also conserved, we can write m 1V1i + m2v2i = myVif + m2V2f A good illustration of such a collision would involve one billiard ball approaching another one that is at rest. Both billiard balls have a mass of m. The first billiard ball has a speed v1; and since the second one is at rest, it has a speed v2 = 0 m/s. In this case, the conservation of kinetic energy can be written as The conservation of momentum will be written as m 1V1i = myVif+ m2V21 Since the masses are all identical, we can write Vii = Vic tv and Vli = Vif + V2f We can rewrite this second equation as Vif = 1i - V2f . By substituting this into the first equation, Vi = (Vi - V2F) + Vz = VH-2v2 Vil + 2Vz After simplifying, 0 = V2f(V1; - V2f) There are two possible solutions: V2f = 0 and V2f = 1j. The first solution V2f = 0 corresponds to no collision at all; particle 2 is still at rest. The second solution, V2f = 1i is the interesting one (there is a collision). When we substitute this value of V2f into the momentum equation, V1; = V1f + V26, we find that V1; = Vif + V1i, and V1f = 0 m/s, so the first mass is brought to rest. To summarize, before the collision the first mass was moving and the second was stationary. After the collision, the first mass came to a stop and the second was moving with the same speed that the first mass had initially. This conclusion is valid only if the two original masses have the same mass and if the collision is perfectly elastic. This is nearly the situation with billiard balls.Since the second mass is initially at rest, we can write the following two sets of equations. equation 1 m1vii = myVif+ m2V2f equation 2 By multiplying both sides of equation 1 by "2" and rearranging, we have m, (Vji - vis ) = m2V25 equation 1a Rearranging equation 2, we have m1 ( V1i - V1f = m2V 2f equation 2a Dividing equation 1a by equation 2a, we have or V1i + V1f = V2f equation 3 Substituting the value for V2f from equation 3 in equation 2 gives m1vi = mivif + m2 ( V 1i + V1f ) or myvi = mivif + m2v 1i + m2 v if m - mz Solving for Vif gives Vit = -Vli m + my By substituting the value for Wif in equation 3 in equation 2, we have m1vli = m1 ( V21 - V1i ) + m2v 2f or miv1i = myv21 - miv1i + m2v2f 2m Solving for V2f gives V2f Vli m + my/ General Statements About the Second Velocities The two equations for the second velocity of my and m, are mm, Ve =v; and Vyg= m, +m, m, + m, 2m, Vi It is interesting to examine these equations for the final velocities of the two objects under three sets of conditions. 1. When the moving object has a greater mass (m4>mj), then vy; is positive, indicating that the moving mass continues in the same direction at a slower speed after the collision. 2. When the moving object has a smaller mass (m{m2), then if is positive, indicating that the moving mass continues in the same direction at a slower speed after the collision. 2. When the moving object has a smaller mass (m,