10. Collisions. Before During After


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1 10. Collisions Use conseation of momentum and enegy and the cente of mass to undestand collisions between two objects. Duing a collision, two o moe objects exet a foce on one anothe fo a shot time: F(t) F(t) Befoe Duing Afte It is not necessay fo the objects to touch duing a collision, e.g. an asteoid flied by the eath is consideed a collision because its path is changed due to the gaitational attaction of the eath. One can still use conseation of momentum and enegy to analyze the collision. Impulse: Duing a collision, the objects exet a foce on one anothe. This foce may be complicated and change with time. Howee, fom Newton's 3d
2 Law, the two objects must exet an equal and opposite foce on one anothe. F(t) t i t Fom Newton's 2nd Law: dp dt = F (t) dp = F (t)dt p f p i = p = ti t f t f t F (t)dt The change in the momentum is defined as the impulse of the collision. Impulse is a ecto quantity. ImpulseLinea Momentum Theoem: In a collision, the impulse on an object is equal to the change in momentum: J = p Conseation of Linea Momentum: In a system of two o moe paticles that ae colliding, the foces that these objects exet on
3 one anothe ae intenal foces. These intenal foces cannot change the momentum of the system. Only an extenal foce can change the momentum. The linea momentum of a closed isolated system is conseed duing a collision of objects within the system. This follows diectly fom Newton's 2nd Law: p dt = p = constant d F ext = 0 Fo a twobody collision, the change in momentum of one object is equal and opposite to the change in the momentum of the othe object: p i = p f p 1i + p 2i = p 1 f p 1 f p 1i = ( p 2 f p 1 = p 2 + p 2 f p 2i )
4 This can also be undestood fom the impulses delieed by the two equal but opposite intenal foces: p 1 = F int dt = J p 2 = ( F int )dt = J p 1 = p 2 Elastic Collision: In an elastic collision, the kinetic enegy of the system is conseed duing the collision. The kinetic enegy of each object will change Conside a collision in which one of the objects is stationay befoe the collision: befoe m m +x afte 1 1 i f f Conseation of momentum fo an isolated system: m 1 1i = m 1 1 f + m 2 2 f m 1 ( 1i 1 f ) = m 2 2 f (1)
5 Conseation of enegy fo an elastic collision: 1 2 m 1 1i = 1 2 m 1 1f m f m 1 2 1i m f = m 2 2 f 2 m 1 ( 1i 1f )( 1i + 1 f ) = m 2 2 f (2) (2) (1): 1i + 1 f = 2 f (3) Substitute into (1): m 1 ( 1i 1f ) = m 2 ( 1i + 1 f ) 1 f = m 1 m 2 1i Substitute into (3): 2 f = 1i + m 1 m 2 1i 2m = 1 m 1 + m 1i 2 Paticle #2 always moes in the positie diection. Paticle #1 will moe in the positie
6 diection if m 1 > m 2 else it will ecoil and moe in the negatie diection. If m 1 = m 2, 1 f = 0 2 f = 1i i.e. paticle #1 comes to est and paticle #2 moes off with oiginal speed of paticle #1. If m 2 >> m 1, 1 f 1i 2 f 2m 1 m 1i << 1i 2 i.e. paticle #1 bounces back in the opposite diection with almost the same speed it had oiginally. The massie paticle #2 moes slowly in the oiginal diection of paticle #1. If m 1 >> m 2, 1f 1i 2 f 2 1i i.e. the massie paticle #1 continues to moe almost as if it did not hit anything, wheeas
7 paticle #2 flies off with twice the initial speed of the massie paticle. We can sole fo the case whee both paticles ae moing befoe the collision using conseation of momentum and enegy: 1 f = m 1 m 2 1i + 2m 2 2i 2 f = 2m 1 1i + m 2 m 1 2i The labeling "1" and "2" is completely abitay. We can change 1 2 eeywhee and get back the same answe. Setting 2i = 0 yields the peious set of equations. Example: A 350g taget glide is at est on a tack, a distance d = 53 cm fom the end of the tack. A 590g pojectile glide appoaches the taget glide with a elocity 1i = 75 cm / s and collides elastically. The taget glide ebounds elastically
8 fom a shot sping at the end of the tack and meets the pojectile glide fo a second time. How fa fom the end of the tack does this second collision occu? d m 2 The pojectile glide will moe with some speed afte the 1st collision towad the sping. The 1st collision gies the taget some elocity to tael to the sping. The taget ebounds elastically fom the sping with the same speed afte 1st collision. Afte 1st collision: 2m 2 f = 1 m 1 + 1i = 94 cm / s m 2 1f = m 1 m 2 m 1 + 1i m 2 1i m 1 = 19 cm / s x = 0 x x = d path of taget path of pojectile
9 t = d + x = d x 2 f 1 f ` x = d 2 f 1f = 34 cm 2 f + 1f Conceptual Question: A popula swingingballs appaatus consists of an aligned ow of identical elastic balls that ae suspended by stings so that they baely touch each othe. When two balls ae lifted fom one end and eleased, they stike the ow and two balls pop out fom the othe end, If instead one ball popped out with twice the elocity of the othe two. This would iolate the conseation of: 1. momentum 2. enegy 3. both of these 4. none  it is possible fo one ball to fly out
10 Inelastic Collision: In an inelastic collision between two objects in an isolated system, kinetic enegy is not conseed, but the linea momentum is conseed. Most collision we obsee eeyday ae inelastic with some loss of kinetic enegy. stating height ending height If we dop a ball fom a height h, it will collide with the eath (gound) and bounce back up. Howee, it will not etun to the oiginal height because some kinetic enegy is lost duing the collision. When two cas collide with one anothe, pats of the cas cumple and bend. Some of the kinetic enegy of the system goes into this defomation, so kinetic enegy is lost duing the collision. This is pat of the design by automatic enginees by haing a "cumple zone" to take enegy out of the collision to potect the die.
11 Completely Inelastic Collision: Two objects stick togethe afte the collision. Afte the collision, the combined objects hae a mass equal to the sum of the two masses and moe with the elocity of the cente of mass. Linea momentum is conseed. Maximum amount of kinetic enegy is lost. p i sys = p f sys m 1 1i + m 2 2i = ( ) = cm = m 1 1 i + m 2 2i Example: A bullet of mass 4.5 g is fied hoizontally into a 2.4kg wooden block at est on a hoizontal
12 suface. The coefficient of kinetic fiction between the block and the suface is The bullet comes to est in the block, which moes 1.8 m. (a) What is the speed of the block immediately afte the bullet comes to est within it, and (b) at what speed is the bullet fied? Befoe collision Afte collision Afte slide b i d x 0 x 0 x 0 = 0 f
13 (a) f = µn = µ(m b + m w )g (m b + m w )a = µ(m b + m w )g a = µg 2 = a(x x 0 ) 0 = i 2 2µgd i = 2µgd = 2.65 m / s (b) Conseation of momentum: m b b + 0 = (m b + m w ) i i b = m b + m w m b = 1400 m / s Conceptual Question: A piece of taffy slams into and sticks to anothe identical piece at est. The momentum
14 of the two pieces stuck togethe afte the collision is the same as it was befoe the collision, but this is not tue of the kinetic enegy which is patly tuned into heat. What pecentage of the kinetic enegy is tuned into heat? 1. 0% 2. 25% 3. 50% 4. 75% 5. moe infomation must be gien Collisions in 2Dimensions: When we studied the motion of objects, we stated by studying motion in one dimension and then we found that eey thing we leaned could be easily applied to 2 o 3 dimensions. The same is tue fo collisions in 2 o 3 dimensions: Linea momentum is each diection is conseed. In elastic collisions, the kinetic enegy of the system is conseed. In totally inelastic collisions, the two objects stick togethe and moe with a common elocity, the elocity of the cente of mass.
15 y 1i F 21 1f 2i F = F f x p sys sys ix = p f x m 1 1i x + m 2 2i x = m 1 1 f x + m 2 2 f x p sys sys iy = p f y m 1 1i y + m 2 2i y = m 1 1 f y + m 2 2 f y Fo elastic collisions: 1 2 m 1 2 1i m 2 2 2i = 1 2 m f m f Example: One ice skate (m 1 ) is skating due east along a fozen lake with a speed 1. Anothe skate (m 2 ) is skating due noth with speed 2. They collide
16 into a big heap and slide acoss the ice togethe. What is thei elocity afte the collision? m 1 1 y f θ x m 2 2
17 p sys sys ix = p f x m 1 1 = ( ) f x f x = m 1 1 p sys sys iy = p f y m 2 2 = ( ) f y f y = m 2 2 θ = tan 1 f y f x = tan 1 m 2 2 m 1 1 f = 2 2 f x + f y = (m 1 1 )2 + (m 2 2 ) 2 Check: If two skates of equal momentum collided: m 1 1 = m 2 2 θ = 45 If skate 2 was stationay:
18 θ = tan 1 (0) = 0 Conceptual Question: (a) The figues below show the position s. time plot fo two bodies and thei cente of mass. The two bodies undego a completely inelastic onedimensional collision while moing along the x axis. Which gaph coesponds to a physically impossible situation? x x x x (a) (b) (c) (d) t t t t (b) The figue below shows seen identical blocks on a fictionless floo. Initially, block a and b ae moing ightwad and block g leftwad, each with the same speed. A seies of elastic collisions occu. Afte the last collision, what ae the speeds and diection of motion of each block? a b c d e f g
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