Question (write down the answers in Pre-lab) Below we consider elastic and completely inelastic collisions of two airtrack gliders colliding with each other (as shown in Fig. 1). Please ignore the effect of friction when making predictions. (1) In a real experiment, the glider masses are measured to be m, = 0.2416 kg and m2 = 0.4262 kg. Glider 1 initially moves toward the positive direction and then collides with Glider 2, which is initially at rest (V2,i = 0). The velocities V1,i, V1,f and V2,f are measured for two types of collision, as shown in the table below. Measured velocities (m/s) Collision type V1.i V1.f V2,f Elastic +0.445 -0.120 +0.315 Completely inelastic +0.355 +0.125 +0.125i. Use the measured initial velocity v1;- to predict the final velocity with sign (hint: Eqs. (7)- (9) might be helpful). Compute the % difference between the measured and predicted |measured valuepredicted valuel x 100%. velocities. The % difference is defined as . |pred1cted valuel ii. Use the measured velocities to compute the initial and final total momenta, initial and final total kinetic energies, as well as their % loss rates. The % loss rate is defined as |final valueinitial valuel 0 W x 100 /o. (2) Given that the collision is elastic and Glider 2 is initially at rest (\"02.1- = 0), please use Eqs. (7) and (8) to explain why i. Glider 2 will be always kicked toward the same direction as Glider 1 comes in (um and vu- have the same sign) . ii. Glider 1 will stop (vlf = 0) if both gliders have the same mass (mjL = m2). iii. Glider 1 will bounce back (171'): and 19111- have opposite sign) if it is less massive than Glider 2 (m1 m2). (3) Given that the collision is completely inelastic and Glider 2 is initially at rest (1221!: = 0), K +K K K' | m lease show that the ener loss rate alwa s satisfies M = 2 hint: find p gY V (Ki.1+Ki.2| m1+m2 ( the kinetic energies in terms of 171,!- and use them to compute the loss rate). Explain how this relation tells that the energy is not conserved upon a completely inelastic collision. (4) Please show that vlf = 1:1]:- and 192,): = 172.1- satisfies both Eqs. (5) and (6) and explain why it is not regarded as a physical solution to an elastic collision problem. The momentum of an object is defined to be its velocity multiplied by its mass. It is a vector quantity that points in the same direction as the velocity, > p = m3. (1) The kinetic energy of an object with mass m and speed v is a scalar and is given by 1 2 K=mv . (2) In any system, to say that a quantity is conserved means that its value does not change over time, even though the system and other associated quantities may change. The law of conserva- tion of momentum therefore says that if there is no net external force acting on a system, the total momentum of the system is constant. External forces are those exerted on any part of the system by an outside object, as opposed to internal forces, which are those that parts of the system might exert on itself. For two objects colliding with each other, conservation of momen- tum is expressed as m' 'l' 321 = 131,;r 'l' 32,)\" (3) Here the subscripts 1 and 2 are labels for objects, and i and f mean initial (before collision) and final (after collision), respectively. This equation says that the total momentum of the two- object system before the collision equals the total momentum after. There is a similar expression for the conservation of mechanical energy. For two objects mov- ing and colliding on a level plane (without the change in gravitational potential energy), it reads Kill + Kilz = Kf,l + Kf'z . (4) \f1 1 1 1 Eml (171,02 + 5771207102 = m1(v1,f)2 + m2(172,f)2- (6) Remember that we will have to keep proper track of signs of the velocities. Collisions are described as either elastic or inelastic, depending on which quantities are con- served in them. Elastic collisions are those where both kinetic energy and momentum are con- served, i.e. both Eqs. (5) and (6) hold. In inelastic collisions, momentum is still conserved but kinetic energy is notsome gets converted into non-mechanical energy such as heat, so only Eq. (5) holds. In other words, the rule to remember is that momentum is conserved in all collisions without net external forces, while kinetic energy is conserved only in elastic collisions. For elastic collisions, if the masses and initial velocities are known, the two objects' final veloc- ities can be determined by solving Eqs. (5) and (6). One gets (m1 m2)vl,i + 21712172,,- . (7) 171 = 'f m1 + m2 (m2 77101721 + 277111711 mi + m2 ' (3) sz = For inelastic collisions, energy conservation [or Eq. (6)] does not apply. In general, momentum conservation [or Eq. (5)] itself cannot determine the two final velocities without more details being given. However, there is a special case called a completely inelastic collision, in which the two objects stick together and move as one object after the collision. The two objects thus have the same final velocity, 171; = 172.}: E 17f. Combining this relation and Eq. (5), one obtains _ m1171;\" + mzvzi v, _ (9) ml +m2 (1) i. Fill in the predicted velocity with sign and % difference between the predicted and measured values. (Express answers to 3 sig. figs.) [8] Collision predicted v1,f % diff in V1,f predicted v2,f % diff in v2,f type Elastic Completely inelastic ii. Fill in the initial value, final value, and % loss. (Express answers to 3 sig. figs.) [6] Momentum (kg x m/s) Kinetic Energy (kg x m2/s2) Collision Initial Final Initial Final type % loss % loss Pi,i + Pz,i P1,f + P2,f K1,i + K2,i K1,f + K2,f Elastic Completely inelastic(2) Explain the four statements about the elastic collision experiment. [8] Answer: (3) Show Atkf2 Ki1-Kizl = 2- for a completely inelastic collision and explain how [Ki,1+Ki,2l mitm2 this relation tells that the energy is not conserved. [4] Answer:(4) Show that V1,f = V1,i and V2,f = V2,i satisfy the conservation equations and explain why it is not a solution to an elastic collision problem. [4]