1. Collisions (and explosions) The conservation of energy is a very powerful law. It is easy to use, mainly because it eliminates time and directions. But because it eliminates time and directions, it does have its limitations. In collisions, the directions of the pieces (both before and after) are important. To attack this problem, we will go back to Newton’s Second Law.

2. Newton’s Second Law We have worked with Newton’s Second Law in the following form:  F x = ma x and  F y = ma y, where acceleration is defined as: a x = dv x /dt and a y = dv y /dt. If mass doesn’t change with motion, which seems to be true, at least at everyday speeds, then we can put the m inside the d (or  ):

3. Momentum  F x = d(mv x )/dt and  F y = d(mv y )/dt. We can re-write this as: (  F x )*dt = d(mv x ) and (  F y )*dt = d(mv y ) The quantity on the right side of the above equations, mv, has its own name: momentum: p = mv Since m is a scalar and v is a vector, p is also a vector.

4. Impulse (  F x )*  t =  (mv x ) and (  F y )*  t =  (mv y ) The quantity on the left side of the above equations also has a name: impulse x = (  F x )*  t and impulse y = (  F x )*  t. When we have collisions, the forces of collision usually happen in very short times, and the details of these collisions are hard to determine.

5. Collisions However, since Newton’s Second Law relates the impulse to the change in momentum: (  F x )*dt = d(p x ) and (  F y )*dt = d(p y ), we can determine the impulse by determining the change in momentum. Further, by determining the time during which the collision took place (  t), we can obtain information about the (average) forces of collision!

6. Collisions There is also another use of Newton’s Second Law for collisions when we use it in combination with Newton’s Third Law (F by 2 on 1 = - F by 1 on 2 )  F x on 1 = F ext on 1 + F by 2 on 1 = d(m 1 v x1 )/dt  F x on 2 = F ext on 2 + F by 1 on 2 = d(m 2 v x2 )/dt. If we add the left sides and add the right sides of the above two equations, we get the forces of collision canceling out!

7. Conservation of Momentum F x ext on 1 + F x ext on 2 =  (p x1 + p x2 ) /  t. If the external forces are small, or if the time of the collision,  t, is small, then we have:  (p x1 + p x2 ) = 0. This can be re-written as: (p x1 + p x2 ) i = (p x1 + p x2 ) f. This is called Conservation of Momentum. This is a vector law, so a similar equation holds for each component of momentum.

8. 1-D examples In one dimensional collision cases, we can apply two laws: Conservation of Energy and Conservation of Momentum (here we assume there are no PE’s that change): (1/2)m 1 v 1i 2 + (1/2)m 2 v 2i 2 = (1/2)m 1 v 1f 2 + (1/2)m 2 v 2f 2 + E lost m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f These are two equations with 7 quantities: m 1, m 2, v 1i, v 2i, v 1f, v 2f, E lost. Hence if we know five, we can solve for the other two.

9. 1-D Collisions We can divide the collisions into three cases: 1. The two objects can bounce off of each other without any E lost. This is called an elastic collision. This means E lost = 0. 2. The two objects can stick to each other. This means v 1f = v 2f. 3. The objects can be deformed but still not stick to one another. No special information is available in these cases.

10. 1-D Collisions In the first two cases, we can predict the final motion (solve for v 1f, v 2f and E lost ) if we know the initial motion (m 1, m 2, v 1i, v 2i ). In the third case, we would have to know something besides the initial motion to solve for the final. When you think about it, this does make sense, since in the third case the results of the collision should depend on the material of the colliding objects!

11. 1-D Example A lead bullet of mass 5 grams collides and sticks inside a block of wood of mass 400 grams. After the collision, the block of wood (with the bullet embedded) moves at a speed of 4 m/s. How fast was the initial speed of the bullet? How much energy was “lost” to denting the block?

12. 1-D example, cont. This is a collision problem, so we have (1/2)m 1 v 1i 2 + (1/2)m 2 v 2i 2 = (1/2)m 1 v 1f 2 + (1/2)m 2 v 2f 2 + E lost m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f We know: m 1 = 5 grams, m 2 = 400 grams, v 2i = 0 m/s (wood block is at rest initially), v 1f = v 2f = 4 m/s (since bullet becomes embedded in block). We are looking for v 1i and E lost.

13. 1-D Example, cont. Using the second equation, we can solve for v 1i : m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f Since v 2i = 0, and v 1f = v 2f = 4 m/s, we have v 1i = (5 grams + 400 grams)*(4 m/s) / 5 grams = 324 m/s. Now we can use the first equation to solve for E lost :

14. 1-D Example, cont. (1/2)m 1 v 1i 2 + (1/2)m 2 v 2i 2 = (1/2)m 1 v 1f 2 + (1/2)m 2 v 2f 2 + E lost E lost = (1/2)*(.005 kg)*(324 m/s) 2 - (1/2)(.005 kg +.400 kg)*(4 m/s) 2 = 259.20 Joules. Note that almost all of the initial energy [total initial energy = (1/2)m 1i v 1i 2 = 262.44 Joules)] went into denting the wood block!

15. 1-D Example modified What would be the speed of the block if the bullet were rubber and bounced off the block instead of denting it and sticking? Would the block be going faster or slower after the collision with the “rubber” bullet as opposed to the “lead” bullet?

16. 1-D Example Modified In this case, we still have a collision, so we have the same equations. But in this case, we know E lost = 0, and v 1i = 324 m/s. We do not know v 1f and v 2f. Thus we still have two equations and need to solve them for two unknowns! Here, however, both unknowns are in both equations. We need to solve them simultaneously.

17. 1-D Example Modified We use the simpler second equation first: m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f to get v 1f = (m 1 v 1i - m 2 v 2f )/m 1 (since v 2i = 0). We use this value for v 1f in the second equation: (1/2)m 1 v 1i 2 + (1/2)m 2 v 2i 2 = (1/2)m 1 v 1f 2 + (1/2)m 2 v 2f 2 + E lost But we know E lost = 0 and v 2i = 0, so we get: (1/2)m 1 v 1i 2 + 0 = (1/2)m 1 [(m 1 v 1i - m 2 v 2f )/m 1 ] 2 + (1/2)m 2 v 2f 2 + 0.

18. 1-D Example Modified (1/2)m 1 v 1i 2 = (1/2)m 1 [(m 1 v 1i - m 2 v 2f )/m 1 ] 2 + (1/2)m 2 v 2f 2. When we multiply out the above equation, we get a quadratic equation for v 2f. This gives two answers: v 2f = 0 m/s or v 2f = 8 m/s. The first answer corresponds to a miss, but that is not what we want. The second answer is what we want.

19. Comparisons Notice that the rubber bullet (v 2f = 8 m/s)gave about twice the kick to the wood block that the lead bullet (v 1f = v 2f = 4 m/s) did! If we think about it using momentum/impulse, the rubber bullet caused the wood block to catch it, and then throw it back! The lead bullet only was caught by the block - it was not thrown back. Hence the block was kicked more by the rubber bullet.

20. Comparisons - a question If we think in terms of conservation of energy, it might at first seem that when the bullet bounces off the target that it carries away kinetic energy that will not be available to the target, while the bullet that becomes embedded does not carry away this energy. This would seem to mean that the target with the embedded bullet should go faster - contrary to our results.

21. Comparisons - explanation What we missed on the last slide was the energy needed to make the hole for the bullet to become embedded (E lost = 259.20 Joules). When we consider this energy, which is not lost in the bouncing bullet case, we see that this energy is substantial, and is enough to make the target with the embedded bullet go slower.

22. 2-D Collisions In two dimensions, we have the scalar Conservation of Energy equation plus the two component equations of Conservation of Momentum. This gives three equations, one more equation than we had with 1-D.. However, we have four more quantities: v 1yi, v 2yi, v 1yf and v 2yf. Usually we know the initial quantities, but this still leaves two more unknowns with only one more equation.

23. 2-D Collisions This does make sense, since there are lots of ways for two objects to collide in 2-D - a head on collision or a variety of glancing collisions! We should need to know something about how the collision happened! A further complication comes in when we realize that we normally express velocities and momenta in polar form (magnitude and angle), but we have to work in rectangular component form.

24. 2-D collisions: rectangular Conservation of Energy: ½m 1 (v 1xi 2 +v 1yi 2 ) + ½m 2 (v 2xi 2 +v 2yi 2 ) = ½m 1 (v 1xf 2 +v 1yf 2 ) + ½m 2 (v 2xf 2 +v 2yf 2 ) + E lost Conservation of momentum in x: m 1 v 1xi + m 2 v 2xi = m 1 v 1xf + m 2 v 2xf Conservation of momentum in y: m 1 v 1yi + m 2 v 2yi = m 1 v 1yf + m 2 v 2yf Quantities: m 1, m 2, v 1xi, v 1yi, v 2xi, v 2yi, v 1xf, v 1yf, v 2xf, v 2yf, E lost

25. 2-D collisions: polar Conservation of Energy: ½m 1 v 1i 2 + ½m 2 v 2i 2 = ½m 1 v 1f 2 + ½m 2 v 2f 2 + E lost Conservation of momentum in x: m 1 v 1i cos(  1i ) + m 2 v 2i cos(  2i ) = m 1 v 1f cos(  1f ) + m 2 v 2f cos(  1f ) Conservation of momentum in y: m 1 v 1i sin(  1i ) + m 2 v 2i sin(  2i ) = m 1 v 1f sin(  1f ) + m 2 v 2f sin(  1f ) Quantities: m 1, m 2, v 1i,  1i, v 2i,  2i, v 1f,  1f, v 2f,  2f, E lost

26. Explosions Explosions can be viewed as collisions run backwards! Instead of E lost, we will need E supplied by the explosion. If you shoot a gun, there is a kick. If you hit a ball, there is a “kick” - if you swing and miss, you are out of balance since you expected a “kick” that didn’t come! From Conservation of Momentum, we can see the reason for the “kick”:

27. Explosions Normally, in an explosion the initial object is in one piece and at rest. After the explosion, one piece goes forward. Conservation of Momentum says the other piece must then go backwards. (If we brace ourselves, we can compensate for that backwards push and not fall over.) 0 + 0 = m 1 v 1f + m 2 v 2f or v 2f = - m 1 v 1f / m 2.

28. Rockets A special case of explosions is when the explosions are controlled and continuous: a rocket and a jet are examples. By throwing the exhaust gases out the back, the engine pushes the body of the plane or rocket forward - by conservation of momentum. But since this is a continuous process (rather than a series of discrete shots), we need the calculus to derive a nice formula.

29. Rockets discrete case Initial: Final M M - ΔMΔM V V - v exhaust V + ΔV

30. Rockets - setup of problem Let’s start with a discrete explosion, and then proceed to the limiting case to get the continuous case. Let: M = mass of rocket before the explosion; V = velocity of rocket before;  M = mass of fuel used in the explosion; v exh = velocity of fuel relative to the rocket;  V = increase in speed of rocket after explosion.

31. Rockets - derivation Conservation of momentum (1-D case): MV = (M-  M)*(V+  V) +  M*(V-v exh ) Note: v exh is a positive quantity; we use the - sign to indicate it is going opposite the direction of the rocket. Multiplying this out gives: MV = MV -  M*V + M*  V -  M*  V +  M*V -  M*v exh Simplifying: 0 = M*  V -  M*  V -  M*v exh

32. Rockets - derivation 0 = M*  V -  M*  V -  M*v exh If we move the - quantities to the left, and divide through by  t, the time for the discrete explosion, we have:  M*(  V/  t) + (  M/  t)*v exh = M*(  V/  t) If we now take the limit as  t  0, (and  M and  V go to zero as well), we have: 0 + (-dM/dt)*v exh = M*dV/dt Note that  M is decreasing M, so  M/  t must become a negative dM/dt in the limit.

33. Rockets - thrust (-dM/dt)*v exh = M*dV/dt Note that the right side is mass times acceleration, so the left side must be the force, or thrust, due to the rocket: F thrust = (-dM/dt)*v exh The thrust of the rocket depends on the exhaust speed of the fuel and the rate at which you use fuel.

34. Rockets - differential equation (-dM/dt)*v exh = M*dV/dt This is a differential equation. To solve it, we can multiply through by dt, then separate the variables (M and V), and then integrate both sides: -dM*v exh = M*dV (-dM/M)*v exh = dV v exh * mo  m -dM/M = vo  v dV v exh * ln[M o /M] = V - V o so finally we have: V = V o + v exh * ln[M o /M]

35. Rockets v f = v o + v exh * ln(m i /m f ) where v exh is the speed with which the rocket or jet engine throws the gases out the back. As the rocket burns its fuel and its speed increases, its mass (including the remaining fuel) goes down. Questions: What does the maximum speed of the rocket depend on? Can the rocket ever go faster than the speed of its exhaust, v exh ?

36. Rockets The Trolley-Sled computer homework program and the Moonlanding computer homework program are examples that show the result of using rockets to cause changes in motion.