Physics 231 Topic 6: Momentum and Collisions Alex Brown October 7 2015 MSU Physics 231 Fall 2015 1
Momentum F = m a Newton s 2nd law F = m v/ t a= v/ t F = m (v final - v inital )/ t Define p = mv p: momentum (kg m/s) F= (p final - p initial )/ t F= p/ t = m v/ t p = m v impulse = mass times change of v The net force acting on an object equals the change in momentum ( p) in time period ( t). Since velocity is a vector, momentum is also a vector, pointing in the same direction as v. MSU Physics 231 Fall 2015 2
Impulse F= p/ t Force=change in (mv) per time period ( t). p=f t The change in momentum equals the force acting on the object times how long you apply the force. Definition: p = Impulse What if the force is not constant within the time period t? F 2 F 1 F 3 p = (F 1 t+ F 2 t+ F 3 t) = area under F vs t plot s s s t MSU Physics 231 Fall 2015 3
car hitting haystack and stoping car hitting wall The change in momentum (impulse) is the same, but the force reaches a much higher value when the car hits a wall! MSU Physics 231 Fall 2015 4
Some examples A tennis player receives a shot approaching him (horizontally) with 50m/s and returns the ball in the opposite direction with 40m/s. The mass of the ball is 0.060 kg. A) What is the impulse delivered by the ball to the racket? B) What is the change in kinetic energy of the ball? A) Impulse=change in momentum ( p). p = m (v final -v initial ) = 0.060(-40-50) = -5.4 kg m/s B) KE i = 75 J KE f = 48 J (no PE!) E nc = KE i -KE f = 27 J (energy lost to heat) MSU Physics 231 Fall 2015 5
Child safety A friend claims that it is safe to go on a car trip with your child without a child seat since he can hold onto your 12kg child even if the car makes a frontal collision (lasting 0.05s and causing the vehicle to stop completely) at v=50 km/h (about 30 miles/h). Is he to be trusted? F = p/ t = m (v f -v i ) / t force = impulse per time period v f = 0 and v i = 50 km/h = 13.9 m/s m = 12kg t=0.05s F = 12(-13.9)/0.05 = -3336 N This force corresponds to lifting a mass of 340 kg!!! MSU Physics 231 Fall 2015 6
Question (assume t is the same for both cases) The velocity change is largest in case: A B The acceleration is largest in case: A B The momentum change is largest in case: A B The impulse is largest in case: A B MSU Physics 231 Fall 2015 7
Conservation of Momentum F 21 t= p 1 = p 1f -p 1i F 12 t= p 2 = p 2f p 2i p 1i p 2i m 1 m 2 Newton s 3rd law: Force of 2 on 1 = F 21 F 12 = - F m 1 m F 12 2 21 F 12 t= -F 21 t p 1f -p 1i = - p 2f + p 2i p 1f p 2f m 1 m 2 Rewrite: p 1i + p 2i = p 1f + p 2f CONSERVATION OF MOMENTUM (only for a closed system) p 1 = m 1 v 1 p 2 = m 2 v 2 MSU Physics 231 Fall 2015 8
question An object is initially at rest in outer space (no gravity). It explodes and breaks into two pieces, not of equal size. The magnitude of the momentum of the larger piece is a) Greater than that of the smaller piece b) Smaller than that of the smaller piece c) The same as that of the smaller piece Initial momentum is zero 0 = p large,final + p small,final So p large,final = - p small,final Magnitude is the same, direction opposite. MSU Physics 231 Fall 2015 9
Quiz An object is initially at rest in outer space (no gravity). It explodes and breaks into three pieces, not of equal size. The momentum of the largest piece: a) Must be larger than that of each of the smaller pieces b) Must be greater than that of each of the smaller pieces c) Must be equal in magnitude but opposite in direction to the sum of the smaller pieces. d) Cannot be determined from the momenta of the smaller pieces The conservation of momentum must hold, even with more than 2 objects. Although the momenta of each piece could be pointing in a different direction, the sum must be equal to zero after the explosion. So, the momentum of one of the pieces must always be opposite in direction, but equal in magnitude to the sum of the momenta of the other two pieces. MSU Physics 231 Fall 2015 10
Moving in space An astronaut (100 kg) is drifting away from the spaceship with v=0.2 m/s. To get back he throws a wrench (2 kg) in the direction away from the ship. With what velocity does he need to throw the wrench to move with v=0.1 m/s towards the ship? (a) 0.1 m/s (b) 0.2 m/s (c) 5 m/s (d) 16 m/s (e) this will never work? Initial momentum: m a v ai + m w v wi =100*0.2+2*0.2 = 20.4 kg m/s After throw: m a v af + m w v wf = 100*(-0.1) + 2*v wf kg m/s Conservation of momentum: m a v ai + m w v wi = m a v af + m w v wf 20.4 = -10 + 2*v wf v wf =15.7 m/s MSU Physics 231 Fall 2015 11
Perfectly inelastic collisions Conservation of momentum: m 1 v 1i +m 2 v 2i =m 1 v 1f +m 2 v 2f After the collision m 1 and m 2 form one new object with mass M = m 1 + m 2 and one velocity v f =v 1f = v 2f m 1 v 1i + m 2 v 2i = v f (m 1 + m 2 ) v f = (m 1 v 1i + m 2 v 2i )/ (m 1 + m 2 ) before after Demo: perfect inelastic collision on airtrack MSU Physics 231 Fall 2015 12
Perfectly inelastic collisions Conservation of momentum: m 1 v 1i +m 2 v 2i =m 1 v 1f +m 2 v 2f After the collision m 1 and m 2 form one new object with mass M = m 1 + m 2 and one velocity v f =v 1f = v 2f m 1 v 1i + m 2 v 2i = v f (m 1 + m 2 ) v f = (m 1 v 1i + m 2 v 2i )/ (m 1 + m 2 ) Let c = m 2 /m 1 then v f = (v 1i + c v 2i )/ (1+c) If m 1 = m 2 (c=1) then v f = (v 1i + v 2i )/ 2 If c=1 and v 2i = 0 then v f = v 1i /2 MSU Physics 231 Fall 2015 13
Perfect inelastic collision: an example 50 m/s 20 m/s 25 m/s A car collides into the back of a truck and their bumpers get stuck. What is the ratio of the mass of the truck and the car? (m truck /m car = m 2 /m 1 =c) What is the fraction of KE lost? v f = [v 1i + cv 2i ] / (1+c) solve for c c= [v 1i -v f ] / [v f v 2i ] = (50-25)/(25-20) = 5 Before collision: KE i = ½m 1 (50) 2 + ½5m 1 (20) 2 After collision: KE f = ½6m 1 (25) 2 Ratio: KE f /KE i = [6*(25) 2 ]/[(50) 2 +5*(20) 2 ] = 0.83 17% of the KE is lost (damage to cars!) MSU Physics 231 Fall 2015 14
Types of collisions Inelastic collisions Momentum is conserved Some energy is lost in the collision: kinetic energy is not conserved Elastic collisions Momentum is conserved No energy is lost in the collision: kinetic energy is conserved Perfectly inelastic: the objects stick together. KE i -KE f = E nc KE i -KE f = 0 MSU Physics 231 Fall 2015 15
Inelastic Solution Conservation of momentum: m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f After the collision m 1 and m 2 form one new object with mass M = m 1 + m 2 and one velocity v f =v 1f = v 2f m 1 v 1i + m 2 v 2i = v f (m 1 + m 2 ) v f = [m 1 v 1i + m 2 v 2i ] / (m 1 +m 2 ) KE i -KE f = E nc With m 2 = c m 1 v f = [v 1i + cv 2i ] / (1+c) Special case m 2 = m 1 (c=1) v f = [v 1i + v 2i ] / 2 MSU Physics 231 Fall 2015 16
Elastic collisions Conservation of momentum: m 1 v 1i +m 2 v 2i =m 1 v 1f +m 2 v 2f Conservation of KE: ½m 1 v 1i2 +½m 2 v 2i2 =½m 1 v 1f2 +½m 2 v 2f 2 Two equations and two unknowns MSU Physics 231 Fall 2015 17
Elastic Solution v 1f = [(m 1 m 2 ) v 1i + 2m 2 v 2i ] / (m 1 +m 2 ) v 2f = [(m 2 m 1 ) v 2i + 2m 1 v 1i ] / (m 1 +m 2 ) KE i -KE f = 0 For m 2 = c m 1 v 1f = [(1-c) v 1i + 2cv 2i ] / (1+c) v 2f = [(c-1) v 2i + 2v 1i ] / (1+c) Special case m 2 = m 1 (c=1) v 1f = v 2i v 2f = v 1i MSU Physics 231 Fall 2015 18
Elastic collisions of equal mass Given m 2 =m 1 and v 1i =0 What are the velocities of m 1 and m 2 after the collision in terms of the initial velocity of m 2 if m 1 is originally at rest? m 1 m 2 For m 2 = c m 1 v 1f = [(1-c) v 1i + 2cv 2i ] / (1+c) v 2f = [(c-1) v 2i + 2v 1i ] / (1+c) m 1 m 2 c = 1 v 1f = v 2i v 2f = v 1i = 0 MSU Physics 231 Fall 2015 19
inelastic collision With m 2 = c m 1 v f = [v 1i + cv 2i ] / (1+c) MSU Physics 231 Fall 2015 20
elastic collisions For m 2 = c m 1 v 1f = [(1-c) v 1i + 2cv 2i ] / (1+c) v 2f = [(c-1) v 2i + 2v 1i ] / (1+c) MSU Physics 231 Fall 2015 21
elastic collisions For m 2 = c m 1 v 1f = [(1-c) v 1i + 2cv 2i ] / (1+c) v 2f = [(c-1) v 2i + 2v 1i ] / (1+c) MSU Physics 231 Fall 2015 22
Quiz In a system with 2 moving objects when a collision occurs between the objects: a) the total kinetic energy is always conserved b) the total momentum is always conserved c) the total kinetic energy and total momentum are always conserved d) neither the kinetic energy nor the momentum is conserved For both inelastic and elastic collisions, the momentum is conserved. MSU Physics 231 Fall 2015 23
Momentum p = mv Impulse p = p f -p i = F t Conservation of momentum (closed system) p 1i + p 2i = p 1f + p 2f Collisions in one dimension given given m 1 and m 2 or c = m 2 /m 1 v 1i and v 2i Inelastic (stick together, KE lost) v f = [v 1i + cv 2i ] / (1+c) Elastic (KE conserved) v 1f = [(1-c) v 1i + 2cv 2i ] / (1+c) v 2f = [(c-1) v 2i + 2v 1i ] / (1+c) MSU Physics 231 Fall 2015 24
A Quiz B Consider a closed system of objects A and B. Object A is thrown at an object B that was originally at rest, as shown in the figure. It is not known if A is heavier or lighter than B. After A collides with B: a) Object A must go left (back in the direction it came from) b) Object A must go right c) Object A could go left or right d) Object B could go left e) The total momentum will be different than before the collision If the collision is perfectly inelastic. A & B stick together and must go right (conservation of momentum). If the collision is elastic and m A <<m B A will go left. If m A >>m B both will go right. So A could go left or right. B can not go left (A would have to go left to, which violates momentum conservation. MSU Physics 231 Fall 2015 25
Ballistic Pendulum M bullet =0.1 kg V bullet =20 m/s M block =1 kg h How high will the block go? There are 2 stages: The collision The Swing of the block The collision The bullet gets stuck in the block (perfect inelastic collision). Use conservation of momentum. m 1 v 1i + m 2 v 2i =v f (m 1 +m 2 ) so: v f =1.8 m/s The swing of the block Use conservation of Mechanical energy. (mgh + ½mv 2 ) start of swing = (mgh + ½mv 2 ) at highest point 0 + ½1.1(1.8) 2 = 1.1*9.81*h + 0 so h=0.165 m Why can t we use Conservation of ME right from the start?? MSU Physics 231 Fall 2015 26
Transporting momentum For elastic collision of equal masses v 2f = 0 v 1f = v 2i 2 1 0 2 1 0 2 1 0 v 2f = 0 v 1f = v 2i v 1f = 0 v 0f = v 1i MSU Physics 231 Fall 2015 27
Quiz The kinetic energy of an object is quadrupled. Its momentum will change by a factor of a) 0 b) 2 c) 4 d) 8 e) 16 MSU Physics 231 Fall 2015 28
Impact of a meteorite Estimate what happens if a 2.8x10 13 kg meteorite collides with earth: a) Is the orbit of earth around the sun changed? b) how much energy is released? Assume: meteorite has same density as earth, the collision is inelastic and the meteorite s v is 10 km/s (relative to earth) A) Earth s mass: m e = 6x10 24 kg m m = 2.8x10 13 kg v e =0 v m =1x10 4 m/s Conservation of momentum: m e v e + m m v m = (m e +m m ) v me =m me v me 0 + (2.8x10 13 )(1x10 4 ) = (6x10 24 )v me so v me =4.7x10-8 m/s (very small) B) Energy=Kinetic energy loss: (½m e v e2 +½m m v m2 )-(½m me v me2 ) 0 + 0.5 (2.8x10 13 ) (1x10 4 ) 2-0.5 (6x10 24 ) (4.7x10-8 ) 2 = 1.4x10 21 J Largest nuclear bomb existing: 100 megaton TNT = 4.2x10 17 J Energy release: 3300 nuclear bombs!!!!! MSU Physics 231 Fall 2015 29
m 1 problem m 1 =5 kg m 2 =10 kg. M 1 slides down and makes an elastic collision with m 2. How high does m 1 go back up? h=5m m 2 ME = mgh + (1/2)mv 2 Step 1. What is the velocity of m 1 just before it hits m 2? Conservation of ME: ME top = ME bottom = m 1 gh + 0 = 0 + 0.5m 1 v 2 So v 1 = 9.9 m/s Step 2. Elastic collision with c = 2, v 1i = 9.9 and v 2i = 0 so v 1f = -3.3 Step 3. m 1 moves back up; use conservation of ME again. ME top = ME bottom = m 1 gh + 0 = 0 + 0.5 m 1 v 1f 2 So h = 0.55 m MSU Physics 231 Fall 2015 30
Playing with blocks m 1 =0.5 kg collision is elastic m 2 =1.0 kg h 1 =2.5 m h 2 =2.0 m A) determine the velocity of the blocks after the collision B) how far away from the bottom of the table does m 2 land MSU Physics 231 Fall 2015 31
Determine the velocity of the blocks after the collision m 1 =0.5 kg collision is elastic m 2 =1.0 kg h 1 =2.5 m h 2 =2.0 m Step 1: determine velocity of m 1 at the bottom of the slide Conservation of ME (mgh+½mv 2 ) top =(mgh+½mv 2 ) bottom 0.5*9.81*2.5 + 0 = 0 + 0.5*0.5*v 2 so: v 1 =7.0 m/s Step 2: Elastic collision equation with c = 2, v 1i = 7 and v 2i = 0 Gives v 1f = -2.33 and v 2f = 4.67 MSU Physics 231 Fall 2015 32
How far away from the table does m 2 land? m 1 =0.5 kg collision is elastic m 2 =1.0 kg h 1 =2.5 m h 2 =2.0 m v 1f =-2.3 m/s v 2f =4.7 m/s This is a parabolic motion with initial horizontal velocity. Horizontal vertical x = v xo t y = y o + v yo t - ½gt 2 x = 4.7 t 0 = 2.0 + 0-0.5*9.81*t 2 = 2.96 m t=0.63 s MSU Physics 231 Fall 2015 33
(A) v y v 0 y at with a = g (B) y v t oy 1 at 2 2 (C) y v t y 1 at 2 2 (D) y 1 2 ( v v ) t oy y (E) y 1 2a ( v 2 y v 2 0 y ) MSU Physics 231 Fall 2015 34
at the top all have v y = 0 (why?) at the top and bottom v x (Q) < v x (N) (why?) at the top v x (P) = v x (R) = 0 (why?) For all (PE+KE) (bottom) = (PE+KE) (top) mv o2 /2 = m(v 2 ox + v oy2 )/2 = mgh + [mv x2 /2] For Q and N get t from (C) MSU Physics 231 Fall 2015 35
a) Speed at X compared to V? b) Velocity at X compared to P? c) Size of total force at R compared to T? d) Size of total force at Q compared to T? MSU Physics 231 Fall 2015 36
a) Speed at X compared to V? X larger than V b) Velocity at X compared to P? same c) Size of total force at R compared to T? T larger than R d) Size of total force at Q compared to T? Q larger than T MSU Physics 231 Fall 2015 37
Remember at the a given time Power = P = F v = m a v MSU Physics 231 Fall 2015 38
question A girl (g) is standing on a plank (p) that is lying on ice (assume frictionless surface between ice and board). She starts to walk with 1 m/s, relative to the plank. What is her speed relative to the ice? Use m g =70 kg m p =100 kg Choose the ice as a frame of reference Initial: both she and the plank are at rest. Final: her velocity relative to the ice: v gp = v g v p v g = v (girl relative to ice) v p = v (plank relative to ice) v p = v g -v gp v gp = v (girl relative to plank) Conservation of momentum: Initial (not moving): =0 Final: m g v g + m p v p = m g v g + m p (v g -v gp ) = 70 v g + 100(v g -1) = 0 Solve for v g = 0.59 m/s MSU Physics 231 Fall 2015 39
Center of Mass/Gravity We often deal with extended objects and need to identify the central (average) location of the mass. This location, also called the center of mass (or gravity) can be found by averaging the x and y positions: MSU Physics 231 Fall 2015 40
Center of Mass/Gravity Given three objects of different mass, find the center of mass of the system. M1: mass = m x = L/2 y = 0 M2: mass = m x = +L/2 y = 0 M3: mass = 2m x = 0 y = L cos(30) M3 M1 L/2 +L/2 M2 MSU Physics 231 Fall 2015 41