1. Let’s look at a bat-ball collision more closely. Before the collision, the ball moves under the influence of gravity and air resistance. The force due to air resistance depends on velocity, and so is time dependent, but over small time periods, the force due to air resistance changes little. Let’s graph these forces. F grav F air force time The impulse delivered to the ball (change in momentum) during a time t is the area under the F vs time curve. tt

2. When the ball is hit by the bat, contact time t is small, and the force of the bat on the ball is huge compared to the forces due to gravity and the air. force time F grav F air tt The impulse delivered to the ball by the bat is typically many hundreds or thousands of times greater than the impulse delivered by gravity and the air. Under these conditions, we can ignore J grav and J air and keep only J bat. F bat

3. What about internal forces, inside the system? By Newton’s third law, every internal force is part of an action-reaction pair. 1 2 F 12 F 21 F 12 =-F 21 so F 12 t=-F 21 t so J 12 =-J 21. The net impulse delivered by this pair of forces is J 12 +J 21 =0. Internal forces do not change the system momentum. The particles may be gyrating wildly around each other, but the system momentum remains constant.

4. It may be that kinetic energy is conserved in a collision. In that case, the collision is “elastic.” You can use K f = K i any time you know a collision is elastic. K f = K i does not appear on your OSE sheet. “So when can I assume a collision is elastic.” Never! (Almost.) (The key word is “assume.”) Demonstration: trust in physics.

5. If the colliding objects are rigid, do not stick together or expend energy deforming, and no energy is lost to heat or sound in the collision, then “elastic” may be a good approximation. Examples of elastic collisions (or close approximations): Billiard balls colliding. Marbles colliding. Carts on air tracks colliding bumper-to-bumper, when the bumpers are made of rubber bands. Atoms, molecules, protons, electrons, etc. colliding (as long as they do not stick together).

6. If kinetic energy is not conserved in a collision, the collision is “inelastic.” Never use K f = K i if you know a collision is inelastic. If the colliding objects deform, stick together, heat up, lose heat or sound energy to their surroundings, etc., the collision is inelastic. Examples of inelastic collisions: Your car and a truck. Chewing gum thrown at the wall. Rain falling in a bucket. Railroad cars being coupled together. A basketball dropped on the floor. You and a truck.

7. Examples (to be worked if time permits) Example: A proton of mass M traveling with a speed V p collides head-on with a helium nucleus of mass 4M at rest. What are the velocities of the proton and helium nucleus after the collision. This is an example of an elastic collision.This is an example of an elastic collision. Why?This is an example of an elastic collision. Why? That’s what section it is in. This is an example of an elastic collision. Why? That’s what section it is in. The book tells you it is. This is an example of an elastic collision. Why? That’s what section it is in. The book tells you it is. Collisions between neutral or like-charged nuclear-type particles are usually elastic. “Head-on” means the collision is not at a glancing angle, which means all motion takes place in one dimension (say, the x-axis). Makes the problem much easier.

8. This is a momentum problem. Back to the litany! Step 1: draw before and after sketch. before M 4M VpVp after M 4M V pf V Hef I chose velocities “positive” (after I put in the x-axis) so I won’t be tricked into putting in an extra – sign later. Step 2: label point masses and draw velocity or momentum vectors (your choice). Already done!

9. before M 4M VpVp after M 4M V pf V Hef Step 3: choose axes, lightly draw in components of any vector not parallel to an axis. xx Step 4: OSE. J ext = P f – P i 0 E f – E i = [W other ] if 0 0 = K f + U gf + U sf – K i – U gi - U si 0 0 00 not really a necessary step, but I thought you should see it

10. Pf = PiPf = Pi K f = K i Step 5: zero out external impulse if appropriate. Already done! before M 4M VpVp after M 4M V pf V Hef xx Continuing from previous slide…

11. Step 6: write out initial and final sums of momenta (not velocities). Zero out where appropriate. This is a combined momentum plus energy problem, but for this step I will just write out the momenta. I’ll put velocities in the next step. before M 4M VpVp after M 4M V pf V Hef xx p pfx + p Hefx = p p + p He,initial 0 Dang! My figure doesn’t justify zeroing out the initial He momentum. Better fix that up right now! V He,initial =0

12. before M 4M VpVp after M 4M V pf V Hef xx V He,initial =0 M V pfx + (4M) V Hefx = M V p Step 7: substitute values based on diagram and solve. K f = K i From conservation of momentum: Conservation of energy: ½M(V pfx ) 2 + ½(4M)(V Hefx ) 2 = ½M(V p ) 2 we want to find V pfx and V Hefx

13. Two equations, two unknowns. How would you solve? M V pfx + (4M) V Hefx = M V p ½M(V pfx ) 2 + ½(4M)(V Hefx ) 2 = ½M(V p ) 2 V pfx + 4V Hefx = V p ----- (1) (V pfx ) 2 + 4(V Hefx ) 2 = (V p ) 2 ----- (2) You would probably solve (1) for V pfx (or V Hefx ) in terms of the other unknown, plug the result into (2), and solve the quadratic. Not impossibly difficult here, but what do you do about the  sign? And what about more complex problems, where the helium initial velocity is nonzero?

14. There’s a mathematical “trick” (actually, technique) that works well here and is the only way to get an easy solution in more complex problems. The technique is used in the physics 31 text to derive an equation which is used a number of times later on. It’s not on your OSE sheet because it is not “fundamental.” Besides, I want you to see this handy technique. V pfx + 4V Hefx = V p ----- (1) (V pfx ) 2 + 4(V Hefx ) 2 = (V p ) 2 ----- (2) To better illustrate the technique, on the next page I will use m in place of the proton mass, and M in place of the helium mass, because in general the masses don’t cancel so conveniently.

15. Group all terms with m on one side, all terms with M on the other side (especially powerful if initial velocity of M is nonzero). mV pfx + MV Hefx = mV p ----- (1) m(V pfx ) 2 + M(V Hefx ) 2 = m(V p ) 2 ----- (2) mV pfx - mV p = -MV Hefx ----- (1) m(V pfx ) 2 - m(V p ) 2 = -M(V Hefx ) 2 ----- (2) The left hand side of equation (2) can be factored mV pfx - mV p = -MV Hefx ----- (1) m(V pfx +V p )(V pfx -V p ) = -M(V Hefx ) 2 ----- (2) Neither side of equation (1) is zero (why?) so divide (2) by (1) to give…

16. mV pfx - mV p = -MV Hefx ----- (1) (V pfx +V p ) = V Hefx ----- (2) Now I can plug V Hefx from (2) back into (1) to get V pfx : mV pfx - mV p = -M (V pfx + V p ) mV pfx - mV p = -MV pfx - MV p mV pfx + MV pfx = mV p - MV p (m+M)V pfx = (m-M)V p V pfx = (m-M) V p / (m+M) You could use this V pfx in (1) or (2) to get V Hefx.

17. mV pfx - mV p = -MV Hefx ----- (1) (V pfx +V p ) = V Hefx ----- (2) I find it easier to solve (2) for V pfx and get V Hefx from (1): m(V Hefx -V p ) - mV p = -MV Hefx mV Hefx - 2mV p = -MV Hefx mV Hefx + MV Hefx = 2mV p (m+M)V Hefx = 2mV p V Hefx = 2mV p / (m+M) Back on slide 7, I said M=4m. In the textbook problem, M=4u and m=1.01u (u=1.66x10 -27 kg).

18. I haven’t boxed my answers, because I am not quite done yet. V pfx = (m-M) V p / (m+M) V Hefx = 2mV p / (m+M) M>m so V pfx is negative, but V Hefx is positive. That means the lightweight proton bounced back to the left after it collided with the helium, and the helium was given a “kick” to the right. That makes sense, so now I put a box around my answers.

19. Sample Inelastic Collision Problems Probably out of time today! Example: A railroad car of mass M moving with a speed V collides with a stationary railroad car of mass M. Calculate how much kinetic energy is transformed to thermal or other forms of energy. Example: A bullet of mass m and speed v is fired into (and sticks in) a hanging block of wood of mass M. The wood is initially at rest. Calculate the height h the wood block rises after the collision.

20. Step 1: draw before and after sketch. Step 2: label point masses and draw velocity or momentum vectors (your choice). Step 3: choose axes, lightly draw in components of any vector not parallel to an axis. Step 4: OSE. Step 5: zero out external impulse if appropriate. Step 6: write out initial and final sums of momenta (not velocities). Zero out where appropriate. Step 7: substitute values based on diagram and solve. Abbreviated Litany for Momentum Problems