Lecture 12 Momentum Energy and Collisions Announcements Assignment

Lecture 12 Momentum, Energy, and Collisions

Announcements Assignment 6 due Wednesday, Oct 5 (11: 59 pm) EXAM: October 13 (through Chapter 9) Look for messages regarding special TA office hours Reminder that formula sheet for Test 2 now on Collab under Midterms Practice problems will be attached to posted slides (this afternoon, I promise) Practice test from last year on Collab under Resources

Reading and Review

Linear Momentum Impulse With no net external force:

A plate drops onto a smooth floor and shatters into three pieces of equal mass. Two of the pieces go off with equal speeds v along the floor, but at right angles to one another. Find the speed and direction of the third piece.

A plate drops onto a smooth floor and shatters into three pieces of equal mass. Two of the pieces go off with equal speeds v along the floor, but at right angles to one another. Find the speed and direction of the third piece. We know that px=0, py = 0 in initial state and no external forces act in the horizontal

An 85 -kg lumberjack stands at one end of a 380 -kg floating log, as shown in the figure. Both the log and the lumberjack are at rest initially. (a) If the lumberjack now trots toward the other end of the log with a speed of 2. 7 m/s relative to the log, what is the lumberjack’s speed relative to the shore? Ignore friction between the log and the water. (b) If the mass of the log had been greater, would the lumberjack’s speed relative to the shore be greater than, less than, or the same as in part (a)? Explain. (c) Check (b) by assuming log has mass of 450 kg.

Rolling in the Rain An open cart rolls along a frictionless track while it is raining. As it rolls, what happens to the speed of the cart as the rain collects in it? (Assume that the rain falls vertically into the box. ) a) speeds up b) maintains constant speed c) slows down d) stops immediately

Rolling in the Rain An open cart rolls along a frictionless track while it is raining. As it rolls, what happens to the speed of the cart as the rain collects in it? (Assume that the rain falls vertically into the box. ) Because the rain falls in vertically, it adds no momentum to the box, thus the box’s momentum is conserved. However, because the mass of the box slowly increases with the added rain, its velocity has to decrease. a) speeds up b) maintains constant speed c) slows down d) stops immediately

Gun Control When a bullet is fired from a gun, the bullet and the gun have equal and opposite momenta. If this is true, then why is the bullet deadly (whereas it is safe to hold the gun while it is fired)? a) it is much sharper than the gun b) it is smaller and can penetrate your body c) it has more kinetic energy than the gun d) it goes a longer distance and gains speed e) it has more momentum than the gun

Gun Control When a bullet is fired from a gun, the bullet and the gun have equal and opposite momenta. If this is true, then why is the bullet deadly (whereas it is safe to hold the gun while it is fired)? a) it is much sharper than the gun b) it is smaller and can penetrate your body c) it has more kinetic energy than the gun d) it goes a longer distance and gains speed e) it has more momentum than the gun Even though it is true that the magnitudes of the momenta of the gun and the bullet are equal, the bullet is less massive and so it has a much higher velocity. Because KE is related to v 2, the bullet has considerably more KE and therefore can do more damage on impact.

Two objects collide. . . and stick mass m No external forces. . . so momentum of system is conserved initial px = mv 0 final px = (2 m)vf mv 0 = (2 m)vf vf = v 0 / 2 A completely inelastic collision: no “bounce back”

What about energy? mass m vf = v 0 / 2 initial final Kinetic energy is lost! KEfinal = 1/2 KEinitial

Inelastic Collisions This is an example of an “inelastic collision” Collision: two objects striking one another “Elastic” collision <=> “things bounce back” Completely inelastic collision: objects stick together afterwards. . . no thing “bounces back” Time of collision is short enough that external forces may be ignored so momentum is conserved Inelastic collision: momentum is conserved but kinetic energy is not

Elastic vs. Inelastic Completely inelastic collision: colliding objects stick together, maximal loss of kinetic energy Inelastic collision: momentum is conserved but kinetic energy is not Elastic collision: momentum and kinetic energy is conserved.

Completely Inelastic Collisions in One Dimension Solving for the final momentum in terms of initial velocities and masses, for a 1 -dimensional, completely inelastic collision between unequal masses: Completely inelastic only (objects stick together, so have same final velocity) Momentum Conservation: KEfinal < KEinitial

Ballistic pendulum: the height h can be found using conservation of mechanical energy after the object is embedded in the block. momentum conservation in inelastic collision vf = m v 0 / (m+M) energy conservation afterwards KE = 1/2 (mv 0)2 / (m+M) PE = (m+M) g h hmax = (mv 0)2 / [2 g (m+M)2]

Binomial expansion If <<1 then (1+ )n ~1+n Ex.

Velocity of the ballistic pendulum Pellet Mass (m): Pendulum Mass (M): Wire length (L): 1. 84 x 10 -3 kg 3. 81 kg 4. 00 m

Crash Cars I If all three collisions below are totally inelastic, which one(s) will bring the car on the left to a complete halt? a) I b) II c) I and II d) II and III e) all three

Crash Cars I If all three collisions below are totally inelastic, which one(s) will bring the car on the left to a complete halt? In case I, the solid wall clearly stops the car. In cases II and III, because ptot = 0 before the collision, then ptot must also be zero after the collision, which means that the car comes to a halt in all three cases. a) I b) II c) I and II d) II and III e) all three

Crash Cars II If all three collisions below are totally inelastic, which one(s) will cause the most damage of lost energy)? (in terms a) I b) II c) III d) II and III e) all three

Crash Cars II If all three collisions below are totally inelastic, which one(s) will cause the most damage of lost energy)? The car on the left loses the same KE in all three cases, but in case III, the car on the right loses the most KE because KE = mv 2 and the car in case III has the largest velocity. (in terms a) I b) II c) III d) II and III e) all three

Inelastic Collisions in 2 Dimensions For collisions in two dimensions, conservation of momentum is applied separately along each axis: Momentum is a vector equation: there is 1 conservation of momentum equation per dimension Energy is not a vector equation: there is only 1 conservation of energy equation

Elastic Collisions In elastic collisions, both kinetic energy and momentum are conserved. One-dimensional elastic collision:

Elastic Collisions in 1 -dimension We have two equations: conservation of momentum conservation of energy and two unknowns (the final speeds). For given case of v 2 i = 0, solving for the final speeds: Note: relative speed is conserved for head-on (1 -D) elastic collision

Limiting cases

Limiting cases

Limiting cases

Toy Pendulum Could two balls recoil and conserve both momentum and energy? Incompatible!

Elastic Collisions I Consider two elastic collisions: 1) a golf ball with speed v hits a stationary bowling ball head-on. 2) a bowling ball with speed v hits a stationary golf ball head-on. In which case does the golf ball have the greater speed after the collision? v at rest a) situation 1 b) situation 2 c) both the same at rest 1 v 2

Elastic Collisions I Consider two elastic collisions: 1) a golf ball with speed v hits a stationary bowling ball head-on. 2) a bowling ball with speed v hits a stationary golf ball head-on. In which case does the golf ball have the greater speed after the collision? Remember that the magnitude of the relative velocity has to be equal before and after the collision! a) situation 1 b) situation 2 c) both the same v 1 In case 1 the bowling ball will almost remain at rest, and the golf ball will bounce back with speed close to v. In case 2 the bowling ball will keep going with speed close to v, hence the golf ball will rebound with speed close to 2 v. v 2 v 2

Elastic Collisions II Carefully place a small rubber ball (mass m) on top a) zero of a much bigger basketball (mass M) and drop these from the same height h so they arrive at the ground with the speed v. What is the velocity of the smaller ball after the basketball hits the ground, reverses direction, and then collides with the small rubber ball? b) v m v v M c) 2 v d) 3 v e) 4 v

Elastic Collisions II Carefully place a small rubber ball (mass m) on top of a much bigger basketball (mass M) and drop these from the same height h so they arrive at the ground with the speed v. What is the velocity of the smaller ball after the basketball hits the ground, reverses direction, and then collides with the small rubber ball? • Remember that relative m velocity has to be equal before v v and after collision! Before the collision, the basketball v bounces up with v and the M v rubber ball is coming down with v, so their relative (a) (b) velocity is – 2 v. After the collision, it therefore has to be +2 v!! a) zero b) v c) 2 v d) 3 v e) 4 v 3 v v (c)

Elastic Collisions in 2 -D Two-dimensional collisions can only be solved if some of the final information is known, such as the final velocity of one object

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A proton collides elastically with another proton that is initially at rest. The incoming proton has an initial speed of 3. 5 x 105 m/s and makes a glancing collision with the second proton. After the collision one proton moves at an angle of 37 o to the original direction of motion, the other recoils at 53 o to that same axis. Find the final speeds of the two protons. v 1 v 0 = 3. 5 x 105 m/s initial 37 o 53 o v 2 final

A proton collides elastically with another proton that is initially at rest. The incoming proton has an initial speed of 3. 5 x 105 m/s and makes a glancing collision with the second proton. After the collision one proton moves at an angle of 37 o to the original direction of motion, the other recoils at 53 o to that same axis. Find the final speeds of the two protons. Momentum conservation: v 1 v 0 = 3. 5 x 105 m/s initial 37 o 53 o v 2 final if we’d been given only 1 angle, would have needed conservation of energy also!

Center of Mass Treat extended mass as a bunch of small masses: In a uniform gravitational field you can treat gravitational force as if it acts at the “Center of Mass”

Center of Mass The center of mass of a system is the point where the system can be balanced in a uniform gravitational field. For two objects: The center of mass is closer to the more massive object.

Center of Mass In general: Symmetry often very useful in determining the Center of Mass

Center of Mass The center of mass need not be within the object

Motion about the Center of Mass The center of mass of a complex or composite object follows a trajectory as if it were a single particle - with mass equal to the complex object, and experiencing a force equal to the sum of all external forces on that complex object

Motion of the center of mass Action/Reaction pairs inside the system cancel out

The total mass multiplied by the acceleration of the center of mass is equal to the net external force The center of mass accelerates just as though it were a point particle of mass M acted on by

Momentum of a composite object

Recoil Speed A cannon sits on a stationary railroad flatcar with a total mass of a) 0 m/s 1000 kg. When a 10 -kg b) 0. 5 m/s to the right cannonball is fired to the left at a c) 1 m/s to the right speed of 50 m/s, what is the d) 20 m/s to the right speed of the center of mass (of e) 50 m/s to the right the flatcar + cannonball)?

Recoil Speed A cannon sits on a stationary railroad flatcar with a total mass of a) 0 m/s 1000 kg. When a 10 -kg b) 0. 5 m/s to the right cannonball is fired to the left at a c) 1 m/s to the right speed of 50 m/s, what is the d) 20 m/s to the right speed of the center of mass (of e) 50 m/s to the right the flatcar + cannonball)? Because the initial momentum of the system was zero, the final total momentum must also be zero, regardless of the release of internal energy, internal forces, etc. If no external forces act, the motion of the center of mass does not change

Recoil Speed II A cannon sits on a stationary a) 0 m/s railroad flatcar with a total mass of b) 0. 5 m/s to the right 1000 kg. When a 10 -kg cannonball is fired to the left at a speed of 50 m/s, what is the recoil speed of the flatcar? c) 1 m/s to the right d) 20 m/s to the right e) 50 m/s to the right

Recoil Speed II A cannon sits on a stationary a) 0 m/s railroad flatcar with a total mass of b) 0. 5 m/s to the right 1000 kg. When a 10 -kg cannonball is fired to the left at a speed of 50 m/s, what is the recoil speed of the flatcar? Because the initial momentum of the system was zero, the final total momentum must also be zero. Thus, the final momenta of the cannonball and the flatcar must be equal and opposite. pcannonball = (10 kg)(50 m/s) = 500 kg-m/s pflatcar = 500 kg-m/s = (1000 kg)(0. 5 m/s) c) 1 m/s to the right d) 20 m/s to the right e) 50 m/s to the right

Center of Mass The disk shown below in (1) clearly has its center of mass at the center. a) higher b) lower Suppose the disk is cut in half and the pieces arranged as shown in (2). Where is the center of mass of (2) as compared to (1) ? c) at the same place d) there is no definable CM in this case (1) X CM (2)

Center of Mass The disk shown below in (1) clearly has its center of mass at the center. a) higher b) lower Suppose the disk is cut in half and the pieces arranged as shown in (2). c) at the same place d) there is no definable CM in this case Where is the center of mass of (2) as compared to (1) ? The CM of each half is closer to the top of the semicircle than the bottom. The CM of the whole system is located at the midpoint of the two semicircle CMs, which is higher than the yellow line. (1) X CM (2) CM