Table of Contents. Pg. # Momentum & Impulse (Bozemanscience Videos) 1 1/11/16

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1 Table of Contents g. # 1 1/11/16 Momentum & Impulse (Bozemanscience Videos) 2 1/13/16 Conservation of Momentum 3 1/19/16 Elastic and Inelastic Collisions 4 1/19/16 Lab 1 Momentum

2 Chapter 6 Work & Energy In physics, work is done by forces on an object over a distance. 1a. W = F d 1b. W = F d Cosθ (Joule, J) 2. Wnet = E ext 3. Kinetic Energy, Ek = 1/2m v 2 (Joule, J) 4. otential Energy, E = m g h (Joule, J) 5. ower, = W t (Watts, W) 5. pring, E = ½kx 2 (Joule, J) Force is parallel to distance Force is at an angle to distance

3 Chapter 7 Momentum & Impulse In physics, work is done by forces on an object over a distance. 1. Impulse, J = F t (N s) 2. Momentum, p = m v (kg m/s) 3. Impulse-Momentum Theorem, F t = p = mvf mvi (kg m/s) 4. Conservation of Momentum, pf = pi (kg m/s) Force F t t f Time t

4 Warm Up (1/19/16) Conservation of Momentum bullet of mass m = 8. g is fired into a block of mass M = 25 g that is initially at rest at the edge of a table of height 5 1. m. The bullet remains in the block, and after the impact the block lands d = 2. m from the bottom of the table. Determine the initial speed of the bullet. By NL, F net = ma = m v = mv; ext t t = p t Thus, = p = p f p i p f = p i h system: m, & M m v b =? M d v initial u final F fr F g F N

5 Warm Up (1/19/16) Conservation of Momentum bullet F of mass m = 8. g is fired into a block net = ma = m v = mv; of mass ext M = 25 t g that t is initially at rest at the edge of a table of height m v 5 1. b =? = p m. MThe bullet t F remains in the block, and after the impact the fr Thus, = p p block lands d = f = p 2. i m from the bottom of the h F table. Determine the initial speed of the bullet. g y NL, = p f p i system: m, & M Figure 6.4 v initial u final ince momentum is conserved, m b v b + Mv B = (m b + M)u v fy 2 = v iy 2 2gh v fy = 2gh t = qr[ 2(9.8m/s 2 )( 1m)] 9.8m/s 2 v m b v b + = (m b fy = v iy gt + M)u v b = (m b + M)u t = v fy g = 2gh m b g =.45 s u = d t = 2m.45s = 4.44 m/s d F N v b = ( )kg 4.44m/s= 143 m/s.8 kg

6 7.2 Conservation of Momentum collision consists of three phases: incoming, interaction, and outgoing.

7 7.2 Conservation of Momentum collision consists of three phases: incoming, interaction, and outgoing. By NL, F net = ma = m v = mv = p ext t t t Thus, = p p f = p i Momentum is Conserved!

8 7.3 Collisions in One Dimension collision consists of three phases: incoming, interaction, and outgoing. erfectly Inelastic Collision Fnet = p; system: ext t = p; system: Elastic Collision Momentum is conserved!

9 Warm Up (1/2/16) Collisions 9-g ball moving at 1m/s collides head-on with a stationary 1-g ball. Determine the speed of each after impact if (a) they stick together (b) they bounce off each other. Before Collision fter Collision v 2 = v 1 u =? m 1 m 2 m 1 m 2 a) Inelastic Collision (balls stick together) F net = ma = m v = mv = p ext t t t = p p f = p i t m 1 v 1 + m 2 v 2 = (m 1 + m 2 )u = p + u = m 1 v 1 (m 1 + m 2 ) u =.9kg(1m/s) (.9 +.1)kg u =.9 m/s

10 Warm Up (1/2/16) Collisions 9-g ball moving at 1m/s collides head-on with a stationary 1-g ball. Determine the speed of each after impact if (a) they stick together (b) they bounce off each other. Before Collision Collision fter Collision b) Elastic Collision (balls bounce off each other) m 1 v 1 v 2 = u 1 =? u 2 =? F net = ma = m v = mv = p ext t t t = p p f = p i t m 1 v 1 + m 2 v 2 = m 1 u 1 + m 2 u 2 = p +

11 Warm Up (1/2/16) Collisions 9-g ball moving at 1m/s collides head-on with a stationary m 1 v 2 = 1-g ball. Determine u 1 =? u 2 = the? speed v 1 of each after impact if (a) they stick together (b) they bounce off each other. ) Elastic Collision (balls bounce off each other) Before Collision F net = ma = m v = mv = ext t t = p p f = p i t = p In an elastic collision, kinetic energy is conserved. KE f = KE i ½m 1 u ½m 2 u 22 = ½m 1 v ½m 2 v 2 2 fter a lot of painful algebra, m 1 v 1 + m 2 v 2 = Collision p t m 1 u 1 + m 2 u 2 u 1 = 2m 2 v 2 + v 1 (m 1 m 2 ) m 1 + m 2 we get: u 2 = 2m 1 v 1 + v 2 (m 2 m 1 ) m 1 + m 2 + fter Collision u 1 =.8m/s = + 1m/s(.8kg).1kg u 2 = 1.8m/s = 2(.9kg)(1m/s) +.1kg

12 (i) m 1 v 1 + m 2 v 2 = u 1 m 1 + u 2 m 2 Conservation of Momentum (ii) ½m 1 u ½m 2 u 22 = ½m 1 v ½m 2 v 2 2 Conservation of KE olving equation i for u 2, we get: (iii) u 2 = v 2 + m 1 (v 1 u 1 ) m 2 Multiplying equation ii, by ½ and rearranging, we get: m 1 (v 1 2 u 1 2 ) = m 2 (u 2 2 v 2 2 ) (iv) m 1 (v 1 u 1 )(v 1 + u 1 ) = m 2 (u 2 v 2 )(u 2 + v 2 ) ubstituting equation iii into equation iv, we get: m 1 (v 1 u 1 )(v 1 + u 1 ) = m 2 v 2 + m 1 (v 1 u 1 ) v 2 (u 2 + v 2 ) m 2 m 1 (v 1 u 1 )(v 1 + u 1 ) = m 2 m 1 (v 1 u 1 )(u 2 + v 2 ) m 2 masses cancel!

13 (v) (v 1 u 1 )(v 1 + u 1 ) = (v 1 u 1 )(u 2 + v 2 ) Note that one possible solution is that v 1 = u 1. In this case note that from equation ii follows that v 2 = u 2. This means that in this case the particles don t interact. Thus, the solution that we want has v 1 u 1, so we may cancel the factor of (v 1 u 1 ) to obtain: (vi) (v 1 + u 1 ) = (u 2 + v 2 ) ubstituting iii into vi, we get: (v 1 + u 1 ) = v 2 + m 1 (v 1 u 1 ) + v 2 olving for u 1, we get: m 2 u 1 ( 1 + m 1 m 2 ) = 2v 2 + v 1 (m 1 m 2 1) u 1 = 2v 2 + v 1 (m 1 m 2 1) ( 1 + m 1 m 2 ) = 2m 2 v 2 + v 1 (m 1 m 2 ) ( m 2 + m 1 )

14 ubstituting the result for u 1 into equation vi we get: (v 1 + u 1 ) = (u 2 + v 2 ) 2m v v 2 + v 1 (m 1 m 2 ) = (u 2 + v 2 ) ( m 2 + m 1 ) olving for u 2, we get: u 2 = v 1 v 2 + 2m 2 v 2 + v 1 (m 1 m 2 ) ( m 2 + m 1 ) u 2 = 2m 1 v 1 + v 2 (m 2 m 1 ) ( m 2 + m 1 )

15 Lab 1 Momentum Questions Under Consideration 1) What happens in a collision? 2) ow could you test the Law of Conservation of Momentum?

16 (a) Warm Up (1/21/16) Ballistic endulum The ballistic pendulum is a device used to measure the speed of a fast-moving projectile such as a bullet. The bullet is fired into a large block of wood suspended from some light wires. The bullet is stopped by the block, and the entire system swings up to a height h. It is possible to obtain the initial speed of the bullet by measuring h and the two masses. s an example of the technique, assume that the mass of the bullet, m 1, is 5. g, the mass of the pendulum, m 2, is 1. kg, and h is 5. cm. Find the initial speed of the bullet, v 1. v 1i f m 1 m 2 v m 1 + m 2 h

17 Warm Up (1/21/16) Ballistic endulum F net = ma = m v = mv = p ext t t t = p p f = p i t m 1 v 1 + m 2 v 2 = (m 1 + m 2 )u = p v 1 = (m 1 + m 2 )u v1 = ( )kg(.99m/s) = 199 m/s.5 kg ince Energy is conserved, E i = E f. u = 2g h = 2(9.8m/s 2 ).5m =.99 m/s Wnet = E ext = E m 1 E i = E f, h =, u = KE i + E i = E f + KE f ½(m 1 +m 2 )u 2 = (m 1 + m 2 )g h v 1i f m 1 m 2 (a) v m 1 + m 2 h

18 FIGURE 7;19 Warm Up (1/22/16) Glancing Collision Billiard ball moving with speed v = 3. m/s in the +x direction strikes an equal-mass ball B initially at rest. The two balls are observed to move off at 45 to the x axis, ball above the x axis and ball B below. That is, θ = 45 and θ B = 45. What are the speeds of the two balls after the collision? pplying the principle of conservation of momentum we get:, at rest! m v + m B v B = m u + m B u B ( m v ) ^ x + ( ) ^ y = (m u Cosθ + m B u B Cosθ B )^x + v B (m u inθ m B u B inθ B )^y B y y θ = 45 B x θ = θ B = u =? v B θ B = 45 v B B =? x u B =?

19 Warm Up (1/22/16) Ballistic endulum ( m v ) ^ x + ( ) ^ y = (m u Cosθ + m B u B Cosθ B )^x + (m u inθ m B u B inθ B )^y Thus, mv = mucosθ + mbubcosθb (muinθ mbubinθb) = v = ucosθ + ubcosθb (uinθ ubinθb) = u = v ubcosθb ubinθb = uinθ v B B FIGURE 7;19 y θ = θ = 45 ub(inθb + CosθB Tanθ ) = vtanθ θ B = v Tanθ = 3m/s Tan45 B inθb + CosθB Tanθ in(-45 ) + Cos(-45 ) Tan45 ub = 2.1 m/s ub = Cosθ u = 3m/s 2.1m/sCos45 Cos45 ubinθb = v ubcosθb inθ = 2.1 m/s Cosθ u =? v B θ B = 45 v B B =? x u B =? The angles must be +45 because we already account for the directions!