3.1 Work 3.2 Energy 3.3 Conservative and nonconservative forces 3.4 Power, energy and momentum conservation 3.5 Linear momentum 3.6 Collisions 3.7 Motion in a gravitational potential CHAPTER 3 WORK AND ENERGY

Part 5 Linear momentum

Linear Momentum & Impulse Newton’s 2 nd Law: Fa F = ma This is the most general statement of Newton’s 2 nd Law pv p ≡ mv pv (p is a vector since v is a vector) So p x = mv x and so on (y and z directions) Definition: p Definition: For a single particle, the momentum p is defined as:

Review: which has more momentum? A bullet, mass = 100 grams, speed = 1000m/s A car, mass=1000kg, speed = 10m/s 100 grams = 0.1kg p=mv = 0.1 kg x 1000m/s = 100 kg.m/s p=mv = 1000 kg x 10m/s = 10,000 kg.m/s

Momentum Conservation Momentum conservation - important principle. A vector expression (P x, P y and P z ).

Momentum Conservation Many problems can be addressed through momentum conservation even if other physical quantities (e.g. mechanical energy) are not conserved

Learning check Consider the bullet & block as a system. After the bullet is shot, there are no external forces acting on the system in the x-direction. Momentum is conserved in the x direction! –P x, i = P x, f –mv = (M+m)V Use momentum conservation to compute relation Between v i and V f ??? v V initialfinal x

Force and Impulse in Boxers:

Impulse: (A variable force applied for a given time) l J reflects momentum transfer F t titi tftf tt J Impulse J = area under this curve ! (Transfer of momentum !) Impulse has units of Newton-seconds

Force and Impulse J l Two different collisions can have the same impulse since J depends only on the momentum transfer, NOT the nature of the collision. tt F t F t tt same area F  t big, F small F  t small, F big

Average Force and Impulse tt F t F t tt F av  t big, F av small F av  t small, F av big F av

Check: Force & Impulse A.heavier B.lighter C.same D.can’t tell Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force F acts on each one for exactly 1 second. Which box has the most momentum after the force acts ? How about Impulse ?? FF light heavy A.heavier B.lighter C.same D.can’t tell

Check:Momentum and Impulse A stunt man jumps from the roof of a tall building, but no injury occurs because the person lands on a large, air-filled bag. Which one of the following statements best describes why no injury occurs? (a) The bag provides the necessary force to stop the person. (b) The bag reduces the impulse to the person. (c) The bag reduces the change in momentum. (d) The bag decreases the amount of time during which the momentum is changing and reduces the average force on the person. (e) The bag increases the amount of time during which the momentum is changing and reduces the average force on the person.

Collisions

Elastic vs. Inelastic Collisions An elastic collision: kinetic energy as well as momentum is conserved before and after the collision. K before = K after –Carts colliding with a spring in between, billiard balls, etc. vvivvi An inelastic collision: kinetic energy is not conserved before and after the collision, but momentum is conserved. K before  K after –Car crashes, collisions where objects stick together, etc.

Exercise: Inelastic collision in 1-D A block of mass M is initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a speed V. In terms of m, M, and V : What is the final energy of the system? –Is kinetic energy conserved? v V initialfinal x

Solution Now consider the kinetic energy of the system before and after: Before: After: So Kinetic energy is NOT conserved! Kinetic energy is NOT conserved! (friction stopped the bullet) However, momentum was conserved, and this was useful.

Test: Inelastic Collision in 1-D M + m v v = ? M m V v = 0 ice (no friction) –What is the initial speed of the M car ? –What is the initial energy of the system? –What is the final energy of the system? –Is kinetic energy conserved?

Check: Momentum Conservation Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface. The ball hitting box 1 bounces back, while the ball hitting box 2 gets stuck. –Which box ends up moving faster? (a) (b) (c) (a) Box 1 (b) Box 2 (c) same 1 2 x V1V1 V2V2

Explanation Since the total external force in the x-direction is zero, momentum is conserved along the x-axis. In both cases the initial momentum is the same (mv of ball). In case 1 the ball has negative momentum after the collision, hence the box must have more positive momentum if the total is to be conserved. The speed of the box in case 1 is biggest! 1 2 x V1V1 V2V2

Prove mv init = MV 1 - mv fin V 1 = (mv init + mv fin ) / M mv init = (M+m)V 2 V 2 = mv init / (M+m) V 1 numerator is bigger and its denominator is smaller than that of V 2. V 1 > V x V1V1 V2V2

Test: Inelastic collision in 2-D Consider a collision in 2-D (cars crashing at a slippery intersection...no friction). Finding angle after collision ? vv1vv1 vv2vv2 V beforeafter m1m1 m2m2 m 1 + m 2

Explanation There are no net external forces acting. –Use momentum conservation for both components. X: y: vv1vv1 vv2vv2 V beforeafter m1m1 m2m2 m 1 + m 2

Solution So we know all about the motion after the collision! V V = (V x,V y ) VxVx VyVy 

Inelastic collision in 2-D... We can see the same thing using vectors: P pp1pp1 pp2pp2 P pp1pp1 pp2pp2 

Explosion (inelastic un-collision) fine the relation bewteen V 1 and V 2 Before the explosion: M m1m1 m2m2 v1v1 v2v2 After the explosion:

Explanation for Explosion... PNo external forces, so P is conserved. P Initially: P = 0 Pvv Finally: P = m 1 v 1 + m 2 v 2 = 0 vv m 1 v 1 = - m 2 v 2 M m1m1 m2m2 vv1vv1 vv2vv2

Learning check V R = ?? V 2 = ?? V 1

Motion in a gravitational potential dr l Work dW g done on an object by gravity in a displacement dr is given by: Fdr rrr dW g = F g. dr = (-GMm / R 2 )r. (dRr + Rd  ) dW g = (-GMm / R 2 ) dR (since r.  = 0, r. r = 1) dr Rd  dR R FFgFFg m M dd

Gravitational potential l Integrate dW g to find the total work done by gravity in a “big” displacement: W g = dW g = (-GMm / R 2 ) dR = GMm (1/R 2 - 1/R 1 ) R1R1 R2R2 R1R1 R2R2 F F g (R 1 ) R1R1 R2R2 M F F g (R 2 ) m

Gravitational potential not on the path taken l Work done depends only on R 1 and R 2, not on the path taken. R1R1 R2R2 m M

Gravitational potential Suppose R 1 = R E and R 2 = R E +  y but we have learned that So: W g = -mg  y R E +  y M m RERE R1R1 R2R2 m M

Potential Energy For any conservative force F we can define a potential energy function U in the following way: –The work done by a conservative force is equal and opposite to the change in the potential energy function. This can be written as: Fr W = F. dr = -  U  Fr  U = U 2 - U 1 = -W = - F. dr  r1r1 r2r2 r1r1 r2r2 U2U2 U1U1

Test : Gravitational Potential Energy We have seen that the work done by gravity near the Earth’s surface when an object of mass m is lifted a distance  y is W g = -mg  y Compute the change in potential energy of this object from work done?? Give potential energy at initial point ??  U = U 1 - U 2 =-W g = mg  y yy m W g = -mg  y j

Solution the change in U near the Earth’s surface is:  U = -W g = mg  y = mg(y 2 -y 1 ). arbitrary constant So U = mg y + U 0 where U 0 is an arbitrary constant. Having an arbitrary constant U 0 is equivalent to saying that we can choose the y location where U = 0 to be anywhere we want to. y1y1 m W g = -mg  y j y2y2

Momentum of an object: p = mv Now: return your home work REVIEW Definitions of impulse and momentum Impulse imparted to object 1 by object 2: I 12 = F 12  t

Home work Thesis 1. The applications of conservation laws 2. Human energy and their applications 3. Power and energy transfer in Biology 4. The shooting technique 5. Particle collisions