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Transcript
Chapter 8 Lecture Chapter 8: Momentum © 2016 Pearson Education, Inc. Goals for Chapter 8 • • • • • Study momentum. Understand conservation of momentum. Study momentum changes during collisions. Add time and study impulse. Understand center of mass and how forces act on the c.o.m. • Apply momentum to rocket propulsion. © 2016 Pearson Education, Inc. Chapter 8: Momentum Conservation Impulse Work Distance, l K = (1/2) m v2 Work-Energy Theorem Energy Conservation p=mv Impulse-Momentum Theorem Momentum Conservation Momentum Conservation © 2016 Pearson Education, Inc. Momentum Can Cause Injury (Concussion) • This is a frame of reference problem just like a passenger in a car. When the brain and skull are moving at the same velocity, there is no problem. If the skull changes abruptly and the brain does not, there is a possibility of an injury. © 2016 Pearson Education, Inc. Definitions Momentum Impulse 𝒎 𝒑 = 𝒎𝒗 [𝒌𝒈 𝒔 ] or [N.s] 𝚫𝒑 = 𝑭𝚫𝒕 =F ( 𝒕𝒇 -𝒕𝒊 )= 𝑱 Force-momentum relation Elastic Collision Inelastic Collision 𝑑 𝑷 𝑑𝒕 =𝒎 𝑑𝒗 𝑑𝒕 = 𝒎𝒂 = 𝑭 𝑲𝟏𝒊 + 𝑲𝟐𝒊 = 𝑲𝟏𝒇 + 𝑲𝟐𝒇 𝑲𝟏𝒊 + 𝑲𝟐𝒊 = 𝑲𝟏𝒇 + 𝑲𝟐𝒇 + 𝑸 Here; 𝑸 is other forms of energy Both elastic and inelastic collisions conserve momentum Elastic Collision Inelastic Collision 𝑷𝟏 + 𝑷𝟐 = 𝑷′𝟏 + 𝑷′𝟐 → 𝒎𝟏 𝒗𝟏 + 𝒎𝟐 𝒗𝟐 = 𝒎′𝟏 𝒗′𝟏 + 𝒎′𝟐 𝒗′𝟐 𝑷𝟏 + 𝑷𝟐 = 𝑷′𝟏 + 𝑷′𝟐 → 𝒎𝟏 𝒗𝟏 + 𝒎𝟐 𝒗𝟐 = 𝒎′𝟏 𝒗′𝟏 + 𝒎′𝟐 𝒗′𝟐 Kinetic energy is not conserved in an inelastic collisions Exploding projectile after first stage burned out © 2016 Pearson Education, Inc. Clicker question A space capsule follows a circular orbit around a planet. A shaped charge blows the capsule into two fragments. As shown in the figure, one fragment falls straight toward the planet's surface. How does the other fragment move right after the explosion? a) It continues on the circular orbit of the original capsule. b) It moves straight away from the planet. c) It moves up and to the right. © 2016 Pearson Education, Inc. Examples of 1D Collisions M m M m Momentum Conservation Elastic Collision Momentum : Kinetic Energy : m1v1 m2 v2 m1v1' m2 v2' 2 2 1 1 1 1 m1v12 m2 v22 m1v1' m2 v2' 2 2 2 2 Momentum Conservation Energy Conservation K 1,i K 2,i K 1, f K 2, f K 1,i K 2,i K 1, f K 2, f Q Loss of energy as thermal and other forms of energy Momentum Conservation Example 2 After collision Before collision (totally inelastic collision) m v1 + m v2 = m v1’ + m v2’ v1’ = v2’ Momentum Conservation Railroad cars, locking up after the collision Find 𝒗′ ; when 𝒗𝟏 =0. 𝒎𝟏 𝒗𝟏 + 𝒎𝟐 𝒗𝟐 = (𝒎𝟏 +𝒎𝟐 )𝒗′ 𝒎𝟏 𝒗𝟏 + 𝟎 = 𝒎𝟏 +𝒎𝟐 𝒗′ 𝒗′ = 𝒗′ = case 𝒎𝟏 = 𝒎𝟐 𝒎𝟏 𝒗𝟏 𝟐𝒎𝟏 𝟏 𝟐 = 𝒗𝟏 Kinetic energy in inelastic collision; 𝟏 𝟏 𝟐 𝒎𝟏 𝒗𝟏 𝟐 + 𝟎 ≥? 𝒎𝟏 +𝒎𝟐 𝒗′ 𝟐 ∴ 𝟏 𝒎𝟏 𝒗𝟏 𝟐 𝟐 > 𝟐 𝟏 𝒎𝟏 𝒗𝟏 𝟐 𝟒 𝒎𝟏 𝒗𝟏 𝒎𝟏 +𝒎𝟐 𝟏 𝒎𝟏 𝒗𝟏 𝟐 𝟐 𝟏 𝟐 Kinetic energy is reduced after the collision How to fire a rifle to reduce recoil What is the recoil of the rifle? 𝒎 𝟏 𝟐 >? (𝟐𝒎𝟏 )( 𝒗𝟏 )𝟐 Given; 𝒎𝑩 = 𝟓𝒈; 𝒗𝑩 = 𝟑𝟎𝟎 and 𝒎𝑹 = 𝟓𝒌𝒈 𝒔 𝒎𝑩 𝒗𝑩 + 𝒎𝑹 𝒗𝑹 = 𝟎 + 𝟎 = 𝒎𝑩 𝒗′𝑩 + 𝒎𝑹 𝒗′𝑹 𝒎𝑩 𝒗′𝑩 𝒎 ∴ 𝒎𝑹 = − = −𝟏. 𝟐 𝒎𝑹 𝒔 Elastic Collision between different mass balls Kinetic energy conservation 𝟏 𝟏 𝟏 𝒎𝑨 𝒗𝒙 𝟐 + 𝟎 = 𝒎𝑨 𝒗𝑨𝒙 𝟐 + 𝒎𝑩 𝒗𝑩𝒙 𝟐 𝟐 𝟐 𝟐 Momentum conservation 𝒎𝑨 𝒗𝒙 = 𝒎𝑨 𝒗𝑨𝒙 + 𝒎𝑩 𝒗𝑩𝒙 Note:single subscript before collision double subscript after collision special case: object B initially at rest 𝒎𝑩 𝒗𝑩𝒙 𝟐 = 𝒎𝑨 𝒗𝒙 𝟐 − 𝒗𝑨𝒙 𝟐 = 𝒎𝑨 𝒗𝒙 + 𝒗𝑨𝒙 𝒗𝒙 − 𝒗𝑨𝒙 𝒎𝑩 𝒗𝑩𝒙 = 𝒎𝑨 𝒗𝒙 − 𝒗𝑨𝒙 Divide equation (1) and (2); 𝒗𝑩𝒙 = 𝒗𝒙 + 𝒗𝑨𝒙 𝒗𝑨𝒙 = 𝒎𝑩 𝒗𝒙 + 𝒗𝑨𝒙 = 𝒎𝑨 𝒗𝒙 − 𝒗𝑨𝒙 (1) (2) 𝒎𝑨 −𝒎𝑩 𝒗 𝒎𝑨 +𝒎𝑩 𝒙 𝒗𝑩𝒙 = 𝟐𝒎𝑨 𝒗 𝒎𝑨 +𝒎𝑩 𝒙 𝒗𝑩𝒙 -𝒗𝑨𝒙 =𝒗𝒙 in general𝒗𝑩𝒇 -𝒗𝑨𝒇 = −(𝒗𝑩𝒊 -𝒗𝑨𝒊 ) 𝒎𝑨 −𝒎𝑩 −𝒎 𝒗𝒙 ∝ 𝑩 𝒗𝒙 𝒎𝑨 +𝒎𝑩 +𝒎𝑩 𝒎𝑨 −𝒎𝑩 𝒎 𝒗𝑨𝒙 = 𝒗𝒙 ∝ 𝑨 𝒗𝒙 𝒎𝑨 +𝒎𝑩 𝒎𝑨 𝑚𝐵 > 𝑚𝐴 ; 𝒗𝑨𝒙 = 𝑚𝐴 > 𝑚𝐵 ; ≈ −𝒗𝒙 and 𝒗𝑩𝒙 ≈ 𝟎 ≈ 𝒗𝒙 when 𝒗𝑩𝒙 = 𝟐𝒎𝑨 𝒗 𝒎𝑨 +𝒎𝑩 𝒙 ≈ 𝟐𝒎𝑨 𝒗 𝒎𝑨 𝒙 ≈ 𝟐𝒗𝒙 For 𝑚𝐴 = 𝑚𝐵 ; 𝒗𝑨𝒙 = 𝟎; 𝒗𝑩𝒙 = 𝒗𝒙 (initial velocity) remember playing billiard ( pool) Elastic Collision between different mass balls Kinetic energy conservation 𝟏 𝟏 𝟏 𝒎𝑨 𝒗𝒙 𝟐 + 𝟎 = 𝒎𝑨 𝒗𝑨𝒙 𝟐 + 𝒎𝑩 𝒗𝑩𝒙 𝟐 𝟐 𝟐 𝟐 Momentum conservation 𝒎𝑨 𝒗𝒙 = 𝒎𝑨 𝒗𝑨𝒙 + 𝒎𝑩 𝒗𝑩𝒙 Object B is at rest before collision 𝒎𝑩 𝒗𝑩𝒙 𝟐 = 𝒎𝑨 𝒗𝒙 𝟐 − 𝒗𝑨𝒙 𝟐 = 𝒎𝑨 𝒗𝒙 + 𝒗𝑨𝒙 𝒗𝒙 − 𝒗𝑨𝒙 𝒎𝑩 𝒗𝑩𝒙 = 𝒎𝑨 𝒗𝒙 − 𝒗𝑨𝒙 (1) (2) Divide equation (1) and (2); 𝒗𝑩𝒙 = 𝒗𝒙 + 𝒗𝑨𝒙 𝒎 −𝒎 𝒗𝑨𝒙 = 𝑨 𝑩 𝒗𝒙 𝒎𝑨 +𝒎𝑩 𝒎𝑩 𝒗𝒙 + 𝒗𝑨𝒙 = 𝒎𝑨 𝒗𝒙 − 𝒗𝑨𝒙 𝒎 −𝒎 −𝒎 𝑚𝐵 > 𝑚𝐴 ; 𝒗𝑨𝒙 = 𝑨 𝑩 𝒗𝒙 ∝ 𝑩 𝒗𝒙 𝒎𝑨 +𝒎𝑩 +𝒎𝑩 𝒎𝑨 −𝒎𝑩 𝒎 𝑚𝐴 > 𝑚𝐵 ; 𝒗𝑨𝒙 = 𝒗𝒙 ∝ 𝑨 𝒗𝒙 𝒎𝑨 +𝒎𝑩 𝒎𝑨 𝒗𝑩𝒙 = 𝟐𝒎𝑨 𝒗 𝒎𝑨 +𝒎𝑩 𝒙 𝒗𝑩𝒙 −𝒗𝑨𝒙 = 𝒗𝒙 ≈ −𝒗𝒙 and 𝒗𝑩𝒙 ≈ 𝟎 ≈ 𝒗𝒙 when 𝒗𝑩𝒙 = 𝟐𝒎𝑨 𝒗 𝒎𝑨 +𝒎𝑩 𝒙 ≈ 𝟐𝒎𝑨 𝒗 𝒎𝑨 𝒙 ≈ 𝟐𝒗𝒙 For 𝑚𝐴 = 𝑚𝐵 ; 𝒗𝑨𝒙 = 𝟎; 𝒗𝑩𝒙 = 𝒗𝒙 (initial velocity) remember playing billiard ( pool) Inelastic Collision Elastic Collision Momentum Conservation Elastic Collision on a air track Elastic Collision on an air track Momentum conservation; 𝒎𝑨 𝒗𝑨𝒊 + 𝒎𝑩 𝒗𝑩𝒊 = 𝒎𝑨 𝒗𝑨𝒇 + 𝒎𝑩 𝒗𝑩𝒇 𝟎. 𝟓𝒌𝒈 ∗ 𝟐. 𝟎 𝒎 𝒔 + 𝟎. 𝟑𝒌𝒈 ∗ −𝟐. 𝟎 𝒎 𝒔 = 𝟎. 𝟓𝒌𝒈 𝒗𝑨𝒇 + 𝟎. 𝟑𝒌𝒈 𝒗𝑩𝒇 = 𝟎. 𝟒𝒌𝒈 Note: Relative velocity changes sign in an elastic collision. 𝒗𝑩𝒇 − 𝒗𝑨𝒇 = − 𝒗𝑩𝒊 − 𝒗𝑨𝒊 = − −𝟐. 𝟎 Now; 𝒎 𝒔 𝒎 𝒎 𝒎 − 𝟐. 𝟎 = 𝟒. 𝟎 𝒔 𝒔 𝒔 𝒎 𝟎. 𝟓 𝒗𝑨𝒇 + 𝟎. 𝟑 𝒗𝑩𝒇 = 𝟎. 𝟒 𝒔 𝒎 𝒎 𝒗𝑩𝒇 − 𝒗𝑨𝒇 = 𝟒. 𝟎 𝒗𝑨𝒇 = 𝒗𝑩𝒇 − 𝟒. 𝟎 𝒔 𝒔 𝟎 + 𝟎. 𝟖𝒗𝑩𝒇 𝒎 = 𝟐. 𝟒 𝒔 𝒗𝑩𝒇 = 𝒗𝑨𝒇 Kinetic energy in elastic collision (energy conservation as expected); 𝟏 𝟏 𝟏 𝒎 𝟐 𝟏 𝟐 𝟐 𝑲𝒊 = 𝒎𝑨 𝒗𝑨𝒊 + 𝒎𝑩 𝒗𝑩𝒊 = 𝟎. 𝟓𝒌𝒈 𝟐 + 𝟎. 𝟑𝒌𝒈 𝟐 𝟐 𝟐 𝒔 𝟐 𝑲𝒇 = 𝟏 𝟐 𝟏 𝟐 𝒎𝑨 𝒗𝑨𝒇 + 𝒎𝑩 𝒗𝑩𝒇 = 𝟏 𝟎. 𝟓𝒌𝒈 −𝟏 𝒎 𝟐 + 𝟏 𝟎. 𝟑𝒌𝒈 𝟐. 𝟒 𝒎 =𝟑 𝟎. 𝟖 𝒔 𝒎 = −𝟏 𝒔 𝒎 −𝟐 𝒔 𝟐 𝒎 𝟐 𝟑 = 𝟏. 𝟔 𝑱 = 𝟏. 𝟔 𝑱 An Explosion Considered Backward in Time – Figure 8.8 • Refer to the worked example on page 228. © 2016 Pearson Education, Inc. Inelastic Collision on an air track Inelastic Collision on an air track a) Find the final velocity of the joint on an air track. 𝒎𝑨 𝒗𝑨𝒊 + 𝒎𝑩 𝒗𝑩𝒊 = (𝒎𝑨 +𝒎𝑩 )𝒗𝒇 𝟎. 𝟓𝒌𝒈 ∗ 𝟐. 𝟎 𝒎 𝒔 + 𝟎. 𝟑𝒌𝒈 ∗ −𝟐. 𝟎 ∴ 𝒗𝒇 = 𝟎. 𝟓 𝒎 𝒔 𝒎 𝒔 = 𝟎. 𝟓𝒌𝒈 + 𝟎. 𝟑𝒌𝒈 𝒗𝒇 (towards the right and together) b) Find the total kinetic energy before the collision. 𝟏 𝟏 𝑲𝑨𝒊 = 𝒎𝑨 𝒗𝑨𝒊 𝟐 = ∗ 𝟎. 𝟓 ∗ 𝟒 = 𝟏. 𝟎𝟏 𝑱 𝟐 𝟐 𝟏 𝟏 𝑲𝑩𝒊 = 𝒎𝑩 𝒗𝑩𝒊 𝟐 = ∗ 𝟎. 𝟑 ∗ 𝟒 = 𝟎. 𝟎𝟔 𝑱 𝟐 𝟐 Total 𝑬𝒌𝒊𝒏 𝒃 before collision is 𝑲𝑨𝒊 +𝑲𝑩𝒊 =1.61 J c) Find the total kinetic energy after the collision 𝟏 𝟏 𝒎 𝑬𝒌𝒊𝒏 𝒂 = 𝑲𝒇 = 𝒎𝑨 + 𝒎𝑩 𝒗𝒇 𝟐 = ∗ 𝟎. 𝟓𝒌𝒈 + 𝟎. 𝟑𝒌𝒈 ∗ 𝟎. 𝟓 𝟐 𝟐 𝒔 𝟐 = 𝟎. 𝟏𝟎 𝑱 d) Compare the kinetic energies. 𝑬𝒌𝒊𝒏 𝒃 𝑬𝒌𝒊𝒏 𝒂 = 𝟏.𝟔𝟏 𝟎.𝟏 ≈ 𝟏𝟔 times larger than after the collision e) Where did this energy go? Extra energies converted to the thermal energy after collision Impulsive Force [Example] an impulsive force on a baseball that is struck with a bat has: Impulsive Force [Note] The “impulse’’ concept is most useful for impulsive forces. Momentum Conservation Very short time Very large magnitude <F> ~ 5000 N & Dt ~ 0.01 s Impulse-Momentum Theorem F (t ) F |J | Dp p f - pi F Dt t f - ti J F ( t f - ti ) p f - pi Momentum Conservation Bend your legs while landing Given; m=70kg and h=3m a) What is the velocity just before landing? 𝐾1 + 𝑈1 = 𝐾2 + 𝑈2 𝐾1 − 𝐾2 = 𝑈2 − 𝑈1 1 1 1 𝑚𝑣1 2 − 𝑚𝑣2 2 = 𝑚𝑣1 2 − 0 = −𝑚𝑔 𝑦 − 𝑦𝑜 = 𝑚𝑔ℎ (𝑣1 is landing velocity) 2 2 2 𝑚 ∴ 𝑣1 = 2𝑔ℎ = 2 ∗ 9.8 ∗ 3 = 7.7 𝑠 What is the impulse during landing? m 𝐽 = 𝐹 ∙ ∆𝑡 = ∆P = Pf − Pi = −70kg ∗ 7.7 = −540N. s s b) What is the average force while landing? Stiff- legged: Total displacement of body during landing is d=1cm ∆𝑡 = 𝑑 𝑣1 = 10−2 7.7 = 1.3𝑥10−3 𝑠 and 𝐹 = 𝐽 ∆𝑡 = 540 1.3𝑥10−3 = 4.15𝑥105 𝑁 Soft (bent) -legged: Total displacement of body during landing is d=0.5m 𝑑 0.5 𝐽 540 ∆𝑡 = = = 0.065 𝑠 and 𝐹 = = = 8.32𝑥103 𝑁 𝑣1 7.7 ∆𝑡 The average force is about 50 times less for the bent landing. 0.065 © 2016 Pearson Education, Inc. Clicker question When you catch a softball, you reduce the impact by letting your hand move backward during the catch. This works mainly because a) you increase the time over which the ball slows. b) you reduce the ball's speed relative to your glove. c) you reduce the impulse delivered to your glove. © 2016 Pearson Education, Inc. (A) Momentum Conservation Ballistic Pendulum Express v and v’ in terms of m, M, g, and h. (A) mv = (m+M) v’ (B) K1+Ug1 = K2+Ug2 2 1 (B) Energy Conservation Momentum Conservation Ballistic Pendulum (cont.) • A bullet of mass m and velocity Vo plows into a block of wood with mass M which is part of a pendulum. – How high, h, does the block of wood go? – Is the collision elastic or inelastic? Two parts: 1-collision (momentum is conserved) 2-from low point (after collision) to high point: conservation of energy 1st part: x : m v 0 (M m) v' mv v' (M m) y : 00 00 2nd part: E bottom E top 1 (M m)(v' ) 2 0 0 (M m)gh 2 1 m 2 v2 2 h (v' ) 2g 2g(m M) 2 Ballistic Pendulum numerical example Obtain 𝑣𝑥 from the height of the ballistic pendulum swing? (Unknowns are 𝒗𝒙 and 𝑽𝒙 ) (Given 𝒎 = 𝟎. 𝟎𝟎𝟓𝒌𝒈 and 𝑴 = 𝟐𝒌𝒈 ) Momentum conservation. (𝒗𝒘 = 𝟎) 𝒎𝒗𝒙 + 𝑴𝒗𝒘 = (𝒎 + 𝑴)𝑽𝒙 𝒎+𝑴 ∴ 𝒗𝒙 = 𝑽𝒙 𝒎 Conservation of energy. 𝟏 𝟐 𝑴 + 𝒎 𝑽𝒙 𝟐 = 𝒎 + 𝑴 𝒈𝒉 𝑽𝒙 = 𝒎 𝟐𝒈𝒉 = 𝟎. 𝟕𝟔𝟕 𝒔 𝒎+𝑴 𝒎 ∴ 𝒗𝒙 = 𝟐𝒈𝒉 = 𝟑𝟎𝟕 𝒎 𝒔 Energy of bullet; 𝑲𝒃𝒖𝒍𝒍𝒆𝒕 = 𝟏 𝒎𝒗𝒙 𝟐 = 𝟐𝟑𝟔𝑱 𝟐 Energy of bullet and block; 𝑲𝒃𝒖𝒍𝒍𝒆𝒕 = 𝟏 𝒎 + 𝑴 𝑽𝒙 𝟐 = 𝟎. 𝟔𝑱 𝟐 A bullet is fired into a wooden block, which slides to a certain distance and then comes to rest due to friction. What is the velocity of the bullet 𝒗𝒙 ? (Given 𝝁 = 𝟎. 𝟐, 𝒎 = 𝟎. 𝟎𝟎𝟓𝒌𝒈, 𝒉 = 𝟎. 𝟐𝟑𝒎 and 𝑴 = 𝟏. 𝟐𝒌𝒈 ) Momentum conservation. (𝒗𝒘 = 𝟎) 𝒎𝒗𝒙 + 𝑴𝒗𝒘 = (𝒎 + 𝑴)𝑽𝒙 𝒎+𝑴 ∴ 𝒗𝒙 = 𝑽𝒙 and 𝑭𝑵 = 𝝁 𝒎 + 𝑴 𝒈 𝒎 Conservation of energy. 𝟏 𝟐 𝑴 + 𝒎 𝑽𝒙 𝟐 = 𝑭𝑵 𝒉 = 𝝁 𝒎 + 𝑴 𝒈𝒉 𝑽𝒙 = 𝟐𝝁𝒈𝒉 𝒎+𝑴 𝒎 ∴ 𝒗𝒙 = 𝟐𝝁𝒈𝒉 = 𝟐𝟐𝟗 𝒎 𝒔 Example 8.8 Accident analysis Conservation of momentum 𝒎 (Given 𝒎𝑨 = 𝟏𝟎𝟎𝟎𝒌𝒈, 𝒎𝑩 = 𝟐𝟎𝟎𝟎𝒌𝒈, 𝒗𝑨𝒙 = 𝟎 and 𝒗𝑩𝒙 = 𝟏𝟎 ) 𝒔 𝒎 𝒔 𝒎 = 𝟏𝟎𝟎𝟎 ∗ 𝟏𝟓 + 𝟎 = 𝟏. 𝟓𝒙𝟏𝟎𝟒 𝒌𝒈 𝒔 𝑷𝒙 = 𝑷𝑨𝒙 + 𝑷𝑩𝒙 = 𝒎𝑨 𝒗𝑨𝒙 + 𝒎𝑩 𝒗𝑩𝒙 = 𝟐𝒙𝟏𝟎𝟒 𝒌𝒈 𝑷𝒚 = 𝑷𝑨𝒚 + 𝑷𝑩𝒚 = 𝒎𝑨 𝒗𝑨𝒚 + 𝒎𝑩 𝒗𝑩𝒚 𝑷 = (𝑷𝒙 )𝟐 +(𝑷𝒚 )𝟐 = 𝟐. 𝟓𝒙𝟏𝟎𝟒 𝒌𝒈 tan 𝜽 = 𝑷𝒚 𝑷𝒙 = 𝟏.𝟓 𝟐.𝟎 𝒎 𝒔 = 𝟎. 𝟕𝟓 and 𝜽 = 𝟑𝟕𝒐 ∴ 𝑷 = 𝒎𝑨 + 𝒎𝑩 𝑽 → 𝑽 = 𝑷 𝒎𝑨 +𝒎𝑩 = 𝟖. 𝟑 𝒎 𝒔 (Velocity of wreckage) Throwing a package overboard 𝒎 (Given: 𝒎𝒑𝒂𝒄𝒌𝒂𝒈𝒆 = 𝟓. 𝟒𝒌𝒈, 𝒗𝒑𝒂𝒄𝒌𝒂𝒈𝒆 = 𝟏𝟎 , 𝒎𝒄𝒉𝒊𝒍𝒅 = 𝟐𝟔𝒌𝒈 and 𝒎𝒃𝒐𝒂𝒕 = 𝟓𝟓𝒌𝒈 ) 𝒔 𝟎 = 𝒎𝒃𝒐𝒂𝒕 + 𝒎𝒄𝒉𝒊𝒍𝒅 𝒗𝒃𝒐𝒂𝒕 + 𝒎𝒑𝒂𝒄𝒌𝒂𝒈𝒆 𝒗𝒑𝒂𝒄𝒌𝒂𝒈𝒆 = 𝟓𝟓 + 𝟐𝟔 ∗ 𝒗𝒃𝒐𝒂𝒕 + 𝟓. 𝟒 ∗ 𝟏𝟎 𝒎 ∴ 𝒗𝒃𝒐𝒂𝒕 = 𝟎. 𝟔𝟔𝟕 𝒔 Find the speed of the blocks after collision. (assume elastic collision). How far up does the smaller block go (rebound height)? (Given 𝜽 = 𝟑𝟎𝒐 , 𝒎 = 𝟐. 𝟐𝒌𝒈, 𝒉 = 𝟑. 𝟔𝒎 and 𝑴 = 𝟕𝒌𝒈 ) 𝒎 𝒔 a) Energy and Momentum conservation (∴ 𝒗𝟐 = 𝟎 ) 𝟏 𝒎𝒗𝟏 𝟐 𝟐 = 𝒎𝒈𝒉 𝒗𝟏 = 𝟐𝒈𝒉 = 𝟐 ∗ 𝟗. 𝟖 ∗ 𝟑. 𝟔 = 𝟖. 𝟒 𝒎 𝒔 (speed of small block before collision) 𝒎𝒗𝟏 + 𝑴𝒗𝟐 = 𝒎𝒗′𝟏 + 𝑴𝒗′𝟐 → 𝟐. 𝟐 ∗ 𝟖. 𝟒 + 𝟕 ∗ 𝟎 = 𝟐. 𝟐 ∗ 𝒗′𝟏 + 𝟕 ∗ 𝒗′𝟐 𝒗𝟏 + 𝒗𝟐 = − 𝒗′𝟏 − 𝒗′𝟐 → 𝟖. 𝟒 − 𝟎 = 𝒗′𝟐 − 𝒗′𝟏 𝒎 𝒎 ∴ 𝒗′𝟐 = 𝟒. 𝟎𝟐 and 𝒗′𝟏 = −𝟒. 𝟑𝟖 𝒔 b) Rebound height 𝟐 ∴ 𝒗′𝟏 = 𝟐𝒈𝒉′ → (−𝟒. 𝟑𝟖)𝟐 = 𝟐 ∗ 𝟗. 𝟖 ∗ 𝒉′ 𝒉′ = 𝟎. 𝟕𝟗 𝒎 and 𝐝= 𝒉′ 𝐬𝐢𝐧 𝜽 = 𝟏. 𝟗𝟔𝒎 𝒔 Find the impulse and the average force on the ball, assuming a collision time ∆𝑡 = 0.01𝑠 = 10𝑚𝑠 a) Given; 𝒗𝒊𝒙 = −𝟐𝟎 𝒎 𝒔 and 𝒗𝒊𝒚 = 𝟎 𝒎 𝒔 𝒗𝒇𝒙 = 𝒗𝒇𝒚 = 𝒗𝒇 cos 𝟒𝟓 = 𝟑𝟎 ∗ 𝟎. 𝟕 = 𝟐𝟏 𝒎 𝒔 𝒎 = 𝟏𝟔. 𝟒 𝒌𝒈 𝒔 𝒎 = 𝟎. 𝟒 𝟐𝟏 − 𝟎 = 𝟖. 𝟒 𝒌𝒈 𝒔 𝑱𝒙 = 𝒎 𝒗𝒇𝒙 + 𝒗𝒊𝒙 = 𝟎. 𝟒 𝟐𝟏 − −𝟐𝟎 𝑱𝒚 = 𝒎 𝒗𝒇𝒚 + 𝒗𝒊𝒚 b) The components of the average force on the ball 𝑱𝒙 𝟏𝟔. 𝟒 𝑭𝒂𝒗,𝒙 = = = 𝟏𝟔𝟒𝟎𝑵 ∆𝒕 𝟏𝟎−𝟐 𝑱𝒚 𝟖. 𝟒 𝑭𝒂𝒗,𝒚 = = −𝟐 = 𝟖𝟒𝟎𝑵 ∆𝒕 𝟏𝟎 𝑭𝒂𝒗 = 𝟏𝟔𝟒𝟎𝟐 + 𝟖𝟒𝟎𝟐 = 𝟏. 𝟖𝒙𝟏𝟎𝟑 𝑵 𝟖𝟒𝟎 ∴ 𝜽 = tan−𝟏 = 𝟐𝟕𝒐 𝟏𝟔𝟒𝟎 Center of Mass (CM) What is the “Center of Mass?” • More importantly “Why do we care?” • This is a special point in space where “it’s as if the object could be replaced by all the mass at that one little point” Center of mass Center of Mass (c.m. or CM) The overall motion of a mechanical system can be described in terms of a special point called “center of mass” of the system: Fsystem M system acm where Fsystem is the vector sum of all the forces exerted on the system. Momentum Conservation © 2016 Pearson Education, Inc. How do you calculate CM? 1. Pick an origin 2. Look at each “piece of mass” and figure out how much mass it has and how far it is (vector displacement) from the origin. Take mass times position 3. Add them all up and divide out by the sum of the masses The center of mass is a displacement vector “relative to some origin” Spelling out the math: m1x1 m2x2 Xcmfor2part icles m1 m2 m1x1 m2x2 m3x3 Xcmfor3part icles m1 m2 m3 m1x1 m2x2 m3x3 M etc... Notethatx is the3- D vector displaceme nt Momentum Conservation CM Position (2D) m3 ycm = 0.50 m X m1 + m2 m1 X xcm = 1.33 m m2 + m3 Total momentum in terms of mass Mvcm ma va mb vb mc vc ... p Motion of center of mass Macm ma aa mb ab mc ac ... Fext Momentum Conservation Collision In a Horizontal Plane – Example 8.5 • Refer to the worked example on page 227. © 2016 Pearson Education, Inc. Energy in an Inelastic Collision – Figure 8.11 • Refer to the worked example on page 230. © 2016 Pearson Education, Inc.
