It is v=I0U1-02)+u2 or 2(01+ v2)(Fig. 10-5). Again we note that VIEW FROMLAB VIEW FROM CAR m1+m2=2m(1+2)/2 BEFORE COLLISION bad Thus, using this principle, we can analyze any kind of collision in which two es of equal mass hit each other and stick. In fact, although we have worked only in one dimension, we can find out a great deal about much more complicated Im AFTER COLLISION collisions by imagining that we are riding by in a car in some oblique direction The principle is the same, but the details get somewhat complicated Fig. 10-5. Tw In order to test experimentally whether an object moving with velocity t olliding with an equal one at rest, forms an object moving with velocity v/2,we perform the follow ent with ou in the trough three equally massive objects, two of which are initially joined to gether with our explosive cylinder device, the third being very near to but slightly eparated from these and provided with a sticky bumper so that it will stick to another object which hits it. Now, a moment after the explosion, we have two objects of mass m moving with equal and opposite velocities v. A moment after 2D+△ m■m n that, one of these collides with the third object and makes an object of mass 2m moving, so we believe, with velocity v/2. How do we test whether it is really /2? r-20-tm mdmh0-g By arranging the initial positions of the masses on the trough so that the distances continues to move with velocity y, should cover twice as much distance in a given D-m m o time as the two which are stuck together(allowing for the small distance travelled by the second object before it collided with the third). The mass m and the mass b Fig. 10-6. An experiment to verify 2m should reach the ends at the same time, and when we try it, we find that they mass m with velocity v striking a do( Fig. 10-6) The next problem that we want to work out is what happens if we have two velocity v/2. different masses. Let us take a mass m and a mass 2m and apply our explosive interaction. What will happen then? If, as a result of the explosion, m moves with velocity t, with what velocity does 2m move? The experiment we have just done may be repeated with zero separation between the second and third masses, and when we try it we get the same result, namely, the reacting masses m and 2m VIEW FROM VIEW FROM attain velocities -v and v/2. Thus the direct reaction between m and 2m gives the same result as the symmetrical reaction between m and m, followed by a collision between m and a third mass m in which they stick together. Furthermore, we find m BEFORE COLLISON m [2m that the masses m and 2m returning from the ends of the trough, with their veloci- ies(nearly)exactly reversed, stop dead if they stick together AF TER COLLISION Now the next question we may ask is this. What will happen if a mass m with velocity v, say, hits and sticks to another mass 2m at rest? This is very easy to Fig. 10-7. Two views of an inelas answer using our principle of Galilean relativity, for we simply watch the collision collision between m and 2m which we have just described from a car moving with velocity -v/2(Fig. 10-7) From the car. the velocities are v-e(car) +U/2=3/2 d n2 Thus we have the answer, i. e. the ratio of velocities before and m [mm 3 to 1: If an object of mass m collides with a stationary object of mass 2m, then the 二 whole thing moves off, stuck together, with a velocity 1/3 as much. The genera rule again is that the sum of the products of the masses and the velocities stays the 0 equals 3m times v/3, so we are gradually building up the theorem nm m of the conservation of momentum, piece by piece. arguments, we can predict the Fig. 10-8. Action and reaction be. result of one against three. two against three, etc, The case of two against three, ween 2m and 3m starting from rest, is shown in Fig. 10-8 Ir e find that the mass of the first object times its velocity, plus the mass of the second object times its velocity, is equal to the total mass of the final object times its velocity. These are all examples, then, of the conservation of