Physics 231 Lecture 14

Transcription

1 Physics 231 Lecture 14 Impulses: forces that last a short time Momentum: p = mv Impulse-Momentum theorem: FΔt = Δp = mδv = m( v f v i ) Momentum conservation: p tot,f p 1,f + p 2,f = p 1,i + p 2,i p tot,i Totally inelastic collisions v f = ( m 1 v 1 +m 2 v 2 ) ( m 1 + m 2 ) Main points of this lect.

2 Loncapa hints vδt vδt Hints: Read The small mass m slides with negligible friction and bounces elastically What does this mean? Mechanical energy is conserved. The height and the speed are connected by mechanical energy conservation. These points are equally spaced in time and give the displacement vectors during these time intervals

3 Example A 50.0-kg student climbs a 5.00-m-long rope and stops at the top. (a) What must her average speed be in order to match the power output of a 200-W light bulb? P = F student v y student climbs with constant velocity so F student =mg v y = P mg = 200W 50kg(9.8m / s 2 ) = 0.41m / s (b) Quiz: How much work does she do? a) 2450 J b) 245 J c) 24.5 J d) 2.45 J W = F student Δy = 50kg i 9.8m / s 2 i5m = 2450J

4 Reading Quiz 1. Impulse is A. a force that is applied at a random time. B. a force that is applied very suddenly. C. the area under the force curve in a force-versus-time graph. D. the interval of time that a force lasts. Slide 9-5

5 Impulse: useful concept for forces that last a very short time There are many processes in which forces last a very short time and are difficult to mathematically describe. Examples are: Kicking, striking batting, dribbling a ball. Various types of explosions, firearms, etc. The typical time dependence of such forces whose actions can be best described by the associated impulse is described below: 0.01s impulse = FaveΔt

6 p = mv Momentum The linear momentum of a particle of mass m is: The change in velocity is related to the change in momentum, i.e. impulse: Δp = p ( v ) f p i = m f v i It is related to the average impulsive force: F ave Δt = m a ave Δt F ave Δt = m ( v f v i ) Δt Δt ( ) F ave Δt = m v f v i F ave Δt = Δ p v f = v F i + ave Δt m

7 Conceptual question about momentum Consider two carts, of masses m and 2m, at rest on an air track. If you push first one cart for 3 s and then the other for the same length of time, exerting equal force on each, the momentum of the light cart is a) four times b) twice c) equal to d) one-half e) one-quarter the momentum of the heavy cart. F ave Δt = Δ p = p f p 0 p 0 = 0 F ave Δt = p f

8 Bouncing balls Assuming each ball has the same mass, which ball experiences the larger impulse? a) the first ball b) the second ball Hint: velocity is a vector. Which ball has the largest change in velocity? Δp = p f p i = m( v f v i ) An elastic collision has twice the impulse as a totally inelastic collisions. vf,elastic = v i v f,inelastic = 0 Δ p elastic = p f p i = m v f v i Δ p inelastic = p f p i = m v i ( ) = 2m v i ( ) = m v i

9 Example A 0.4 kg ball is dropped from rest at a point 1.5 m above the floor. The h m ball rebounds straight upward to a height of 0.8m. What is the magnitude and direction of the impulse applied to the ball by the floor? h f 0.8 m If the ball is in contact with the floor for 0.01 seconds, what is the m 0.4 kg impulse force? Δ p = p " after p before = m ( v after v before ); velocities just before and after the collision. Choose 0 the gravitation potential 0 energy to vanish at y=0 1 2 mv 2 + mgh 0 0 = 1 2 mv 2 + mg i 0 before 1 2 mv 2 = mgh before 0 v before = 2gh 0 = 2( 9.8) ( 1.5)m / s = 5.4m / s mv mg i 0 = 1 after 2 mv mg i h f f mv 2 mgh v 2gh 2 after = f after = f = ( 9.8 )( 0.8 ) m / s = 3.96m / s 2 Δ p = m ( v after v before ) = 0.4kg 3.96m / s ( 5.4m / s) = 3.75kg m / s upwards Δp 3.75kg m/s F= = = 375N upwards Δt 0.01s

10 Quiz Jack swings at a 0.2 kg ball that is moving west with a velocity of 40 m/s and hits a line drive. The leaves his bat with a velocity of 40 m/s due east. Assuming the ball is in contact with the bat for s, what is the average impulse force of the bat on the ball? a) 800N east b) 1600 N east c) 1600 N west d) 800 N west m Δt v 0 v f 0.2 kg 0.01 s -40 m/s 40 m/s choose east to be positive ( [ ]) Δ p= 0.2kg 40m/s 40m/s = 16kg m/s east F Δp 16kg m/s = = = 1600N east Δt 0.01s

11 Reading Quiz 2. The total momentum of a system is conserved A. always. B. if no external forces act on the system. C. if no internal forces act on the system. D. never; momentum is only approximately conserved. Slide 9-7

12 Conservation of linear momentum p2, f + p1, f = p2, o p tot, f = p tot, o + p This applies to collision of objects that interact with each other but whose interactions with the rest of the world can be neglected. This the definition of an isolated system. As an example, one can consider a collision between two hockey pucks (one larger and the other smaller) that are sliding without friction on a frictionless ice surface. Proof: From Newton s 3 d law: F 21 = F 12 at all times. Thus: F 21 = F 12 F 21 Δt = F 12 Δt 1, o but F 21 Δt = Δ p 2 and F 12 Δt = Δ p 1 Δ p 2 = Δ p 1 p 2,f p 2,o = p 1,o p 1,f so : p 2, f + p 1, f = p 2,o + p 1,o p tot, f = p tot,o If there are external forces like gravity, momentum may not be conserved. In such cases: F extδt = ptot, f ptot, i

13 Conceptual question Which of these systems are isolated? a) While slipping on a patch of ice (µ k =0), a car collides totally inelastically with another car. System: both cars b) Same situation as in a). System: the slipping car c) A single car slips on a patch of ice. System: car d) A car makes an emergency stop on a road. System: car e) A ball drops to Earth. System: ball f) A billiard ball collides elastically with another billiard ball on a pool table. System: both balls

14 Principles of collisions If there are no external forces, the total momentum is always conserved during a collision: p p + p = p + p tot, f 1, f 2, f 1, i 2, i ptot, i In such collisions, however, the mechanical energy may or may not be conserved. We have two important limits: Totally inelastic collisions where the two objects stick together after the collision. Here, largest energy loss possible for an isolated system occurs. Totally elastic collisions where the two objects bounce off each other and the mechanical energy is the same after the collisions as it is before the collision. Inelastic collisions can occur in which the objects do not stick together. The energy loss in such collisions is less than what occurs in totally inelastic collisions where the object do stick together.

15 Reading Quiz 3. In an inelastic collision, A. impulse is conserved. B. momentum is conserved. C. force is conserved. D. energy is conserved. E. elasticity is conserved. Slide 9-9

16 Totally inelastic collisions In isolated systems (systems without external forces) momentum is conserved. In totally inelastic collisions, the particles stick together after the collision. p tot,0 = m v1,0 1 + m v2,0 2 = m vf 1 + m vf 2 = ( m 1 + m 2 ) v f v f = m v 1 1,0 + m v2,0 2 m 1 + m 2 Example: A 40 kg skater, sliding to the right without friction with a velocity of 1.5 m/s, suffers a head on collisions with a 30 kg skater who is initially at rest. After the collisions the two cling together. What is the final speed? m a) 0.17 m/s m 1 40 kg 1 v b) 0.23 m/s f= v1,0 m1+ m m kg c) 0.42 m/s d) 0.86 m/s 40kg v 1,0 1.5 m/s ( ) = 1.5m / s 0.86m / s 40kg + 30kg = v 2,0 0 v f?