Physics 201, Lecture 15

Transcription

1 Physics 0, Lecture 5 Today s Topics q More on Linear Moentu And Collisions Elastic and Perfect Inelastic Collision (D) Two Diensional Elastic Collisions Exercise: Billiards Board Explosion q Multi-Particle Syste and Center of Mass Ø Hope you ve previewed Chapter 9.

2 Review: Linear Moentu and Moentu Conservation q Linear Moentu p p p p... = v v v... = p j q Ipulse-Moentu theore Δ p = p f p i = q Moentu Conservation: F (t)dt I (ipulse) ext p f = p, if F i ext = 0

3 Review: Collisions q Collision: An event in which two particles coe close and interact with each other by force. Moentu is conserved in collision: P f =P i (Per Ipulse approxiation) Kinetic Energy of the syste ay or ay not be conserved: Elastic: KE f = KE i Inelastic: KE f KE i Two extree cases: Elastic and Perfectly Inelastic.

4 Review: -Dientional Elastic Collision q Take v i =0: If << : v f = -v i, v f = v i =0 (think of a tennis ball hitting ground) If >> : v f = v i, v f = v i If = : v f = 0, v f = v i (deo last Thursday) i i f i i f v v v v v v = =

5 Review: -Dientional Perfectly Inelastic Collision q Perfectly inelastic collision: After collision, two particles have sae velocity v f. q Moentu in x direction: P i =P f v i v i = v f v f à v f = ( v i v i )/ ( ) q Question: Is kinetic energy the sae before and after? Before: KE i = ½ v i ½ v i After: KE f = ½ v f ½ v f à KE f -KE i = - ½ /( ) (v i -v i ) <0!!! Quizzes: What is the work done in collision? Where is the lost energy?

6 Two-Diensional Elastic Collision q Collision can be -D, the sae approach as in D works. v i v f y v i v f x Ø Moentu conservation P i =P f (P ix =P fx, P iy =P fy ) v xi v xi = v xf v xf v yi v yi = v yf v yf Ø Elastic KE i =KE f : ½ v i ½ v i = ½ v f ½ v f Ø Three equations and four unknowns (v fx, v fy, v fx, v fy ) requires one ore assuption to get full solution. Ø The rest is algebra!

7 Glancing Collision y x Target is at rest q Elastic KE i =KE f : ½ v i = ½ v f ½ v f Note: v = v x v y q Moentu conservation P i =P f (P ix =P fx, P iy =P fy ) v xi = v xf v xf v yi = v yf v yf q Three equations and four unknowns (v fx, v fy, v fx, v fy ) Again, requires one extra given condition to get full solution.

8 Exercise: Billiards Board q Find the angle θ of the cue ball after collision. (Assuing elastic collision, and all asses equal) Solution: Elastic: KE i = KE f à ½ v i = ½ v f ½ v f Moentu Conservation p i =p f x: v i = v f cosθ v f cos5 o à y: 0 = -v f sinθ v f sin5 o Solve (exercise after class): cos(θ5 o ) =0 θ5 o =90 o θ=55 o In general: The two balls always akes 90 o after collision!

9 Another Trick for Billiard Board q Show that if elastic collision, and in the liit that table ass M is uch larger than ball ass, θ f = θ i. Keys to Solution: M>> table does not ove. All kinetic energy carried by ball. Elastic: KE f = KE i à v f = v i v fx v fy = v ix v iy Noral force no force in x direction à v fx = v ix è v fy = v iy i.e. v fy = - v iy Trigonoetry: tanθ i = v ix /v iy, tanθ f = v fx /v fy à θ f = θ i

10 Explosion q Explosion: A single object, often at rest, breaks into ultiple oving pieces within a very short period of tie v 4 v 4 M Before Explosion v v After Total oentu is conserved (Ipulse approxiation) p f = p i (note the vector for!) Kinetic Energy is not conserved! Before: v=0 KE i =0 After: KE f = ½ v ½ v ½ v... = Σ ½ j v j >0

11 Quick Quiz before v=0 M after A B Which of these is a possible after state? A B both

12 Quick Quiz before M v=0 after A B Which of these is a possible after state? A B both

13 Exercise: a Siple Explosion q Find v after the string is cut. (M=.00Kg, ignore all frictions) q Solution: p i = 0, p f = Mv M(.00) = p i = 0 v = /s q Energy consideration: before: KE i = 0 after: KE f = ½ M(6.00) ½ M(.00) = 4 J Ø Quiz: Where does this 4J coe fro? Ø Answer: fro energy initially stored in the spring For a real bob, the energy coes fro cheical energy in TNT

14 (After Class) Conceptual Exercise q A gun of ass M gun is firing a bullet of ass M bullet. How does the recoil of the gun depend on the ass of the bullet? q Answer/Solution: This is not an easy quiz at all! It requires a full solution. Ø Proble setting: v Gun v Bullet before all at rest: p i = 0, KE i = 0 after p f = M Gun v Gun M Bullet v Bullet =0 KE f = ½ M Gun v Gun ½ M Bullet v Bullet = E Ø Solve: v Gun ~ (EM Bullet ) ½ /M Gun if M Bullet saller, v Gun saller Energy fro gun powder if M Gun >> M Bullet

15 Multi-Particle Syste and Center of Mass For a ulti-particle syste:,,,... at r, r, r,... one can define: Ø Total ass: M = Σ j =... Ø Center of Mass (CM) position: r r r... r CM M Ø CM Velocity and Acceleration v CM d r CM dt a CM d v CM dt = v v v... M = a a a... M r r CM CM r r Ø Now think of CM as a virtual particle, it has M, r, v, a

16 Exercise: Find Center of Mass q Find the CM for these object syste. (all asses sae) q Soe exaples of CM 0 0 L L x x x x CM = = 0 0 L L y y y y CM = =

17 Quick Quiz: CM Location q For the base ball bat below, which point is closer to the center-of-ass a. A b. B c. C A B C

18 Quick Quiz: Dividing at CM q A baseball bat of unifor density is cut at the location of its center of ass as shown below. Which piece has the saller ass? a. The left piece b. The right piece c. Both pieces have the sae ass

19 Quick Quiz: v CM and Moentu q It is known that at a particular oent, the total oentu of a ulti-particle syste is zero. Which of the following stateents is true? A: The syste s total kinetic energy is zero B: The total external force on the syste is zero C: The center of ass velocity of the syste is zero D: Both B and C above are correct E: None of above is correct v CM d r CM dt = v v v... M = p j M = p M

20 Dynaics of Center of Mass q Ipulse-Moentu Theore: F ext = d p dt F ext = d p dt = dv dv dv... dt = M d v CM dt = M a CM The otion of CM follows Newton s nd Law if only external forces are considered!

21 Motion of CM CM follows projectile trajectory! (if gravitation is the only external force)