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CHAPTER 7 MOMENTUM AND COLLISIONS MOMENTUM Momentum is the product of the mass and velocity of a body. Momentum is a vector quantity that has the same direction as the velocity of the object. MOMENTUM p = mv m = mass (kg) v = velocity (m/s) p = momentum (kg•m/s) PROBLEM 1 A 2250 kg pickup truck has a velocity of 25 m/s to the east. What is the momentum of the truck? m = 2250 kg v = 25 m/s p = mv = 2250(25) = 56250 kg•m/s East IMPULSE-MOMENTUM THEOREM Impulse is the product of the average force and the time interval during which the force is exerted. F = ma Δv F=m Δt F×Δt = m×Δv Impulse = change in momentum Applications of Impulse Momentum Theorem Used to determine stopping distances and safe following distances of cars and trucks. Used in designing safety equipment that reduces force exerted on the object during collisions. Example-nets and giant air mattresses for fire-fighters. Also used in sports equipment and games. PROBLEM 3 A baseball of mass 0.14 kg is moving at 35 m/s. A) Find the momentum of the baseball. B) Find the velocity of a bowling ball, mass 7.26 kg, if its momentum is the same as the baseball. m = 0.14 kg v = 35 m/s p = mv = 0.14(35) = 4.9 kg•m/s p 4.9 kg m / s m 7.26 kg p 4.9 v m 7.26 0.67 m / s PROBLEM 6 The brakes exert a 640 N force on a car weighing 15689 N and moving at 20 m/s. The car finally stops. A) What is the car’s mass? B) What is the car’s initial momentum? C) What is the change in the car’s momentum? D) How long does the braking force act on the car to bring it to a halt? F 640 N W 15689 N t ? vi 20 m / s vf 0 m / s m? pi ? p ? t ? W 15689 N W 15689 m g 9.8 vi 20 m / s 1600.92 kg m 1600.92 kg pi mvi 1600.92 20 32018.4 kg m / s pi 32018 kg m/s pf 0 kg m/s Δp pf pi 32018.4 kg m/s F t p 32018.4 kg m/s t 640 N 50 s CONSERVATION OF MOMENTUM States that the momentum of any closed, isolated system does not change. pinitial pfinal Isolated System has no net force acting on it. Closed System is a system in which objects neither enter nor leave . PROBLEM 11 Glider A of mass 0.355 kg moves along a frictionless air track with a velocity of 0.095 m/s. It collides with a glider B of mass 0.710 kg moving in the same direction at a speed of 0.045 m/s. After collision glider A continues in the same direction with a velocity of 0.035 m/s. What is the velocity of glider B after collision? Glider A m A = 0.355 kg Glider B m B = 0.710 kg v Ai = 0.095 m/s v Bi = 0.045 m/s v Af = 0.035 m/s v Bf = ? m/s Conservation of momentum means: pinitial = pfinal p Ai + pBi = pAf + pBf m A v Ai + mB v Bi = mA v Af + mB vBf m A v Ai + m B v Bi = m A v Af + m B v Bf 0.355(0.095) + (0.710)(0.045) = 0.355(0.035) + (0.710)v Bf 0.033725 + 0.03195 = 0.012425 + (0.710)vBf 0.065675 - 0.012425 = (0.710)vBf 0.05325 = (0.710)vBf v Bf = 0.075 m/s INTERNAL FORCES Internal forces are forces between objects within a system. Example: If we consider a single car as our system, forces are exerted on objects within the car during a collision (i.e. a crash dummy) INTERNAL FORCES EXTERNAL FORCES External force is a force exerted by an object outside the system. Example: Our single car from the previous example is still considered our system. If the car collides with a tree, then the force the tree exerts on the car is an external force. EXTERNAL FORCES PROBLEM 17 Two campers dock a canoe. One camper steps on the dock. This camper has a mass of 80 kg and moves forward at 4 m/s. With what speed and direction do the canoe and the other camper move if their combined mass is 110 kg? Camper m a = 80 kg v ai = 0 m/s Camper and Canoe v af = 4 m/s m b = 110 kg v bi = 0 m/s v bf = ? m/s p ai + p bi = paf + p bf 0 = m a v af + m b v bf v bf -m a v af = mb v bf = -2.9m/s Types of Collisions – objects are apart after the collision ELASTIC – objects “stick” together after the collision INELASTIC COLLISIONS INELASTIC Momentum is conserved Some KE is changed into other forms TOTALLY ELASTIC Momentum is conserved. KE is conserved. ELASTIC COLLISIONS http://www.youtube.com/watch?NR=1&v=SBesU12g58I INELASTIC COLLISIONS http://www.youtube.com/watch?v=wFoPawE0LxA SKATER/MEDICINE BALL EXAMPLE A 15 kg medicine ball is thrown at a velocity of 20 km/hr to a 60 kg skater who is at rest on ice. The skater catches the ball and slides with the ball across the ice. Determine the velocity of the skater and the ball after the collision. The collision occurs between a skater and a medicine ball. Before the collision, the ball has momentum and the person does not. The collision causes the ball to lose momentum and the skater to gain momentum. After the collision, the ball and the skater travel with the same velocity ("v") across the ice. SKATER/MEDICINE BALL EX Before Collision Skater 0 Medicine ball 300 Total 300 After Collision 60 * v 15 * v 300 p before = pafter 60v 15v 300 75v 300 v 4 km/hr GRANNY/RALPH EXAMPLE GRANNY/RALPH EXAMPLE Before Collision After Collision Granny 80 * 6 = 480 80 * v Ralph 0 40 * v Total 480 480 p before = pafter 80v 40v 480 120v 480 v 4m/s CAR/TRUCK EXAMPLE CAR/TRUCK EXAMPLE Before Collision Truck Car Total After Collision 3000 * 10 = 30 000 3000 * v 0 1000 * 15 = 15 000 30 000 30 000 pbefore = pafter 30000 = 3000v +15000 15000 = 3000v v = 5m/s Elastic Elastic Elastic Above is a representation of 3 bullets with equal mass running into 3 blocks of wood with equal mass. The first bullet passes through the block and maintains much of its original momentum As a result, very little momentum gets transferred to the block. The second bullet, expands as it enters the block of wood which prevents it from passing all the way through it. As a result, most of its momentum gets transferred to the block. (This is an inelastic collision.) The third bullet (a rubber bullet) bounces off the block transferring all of it's own momentum and then borrowing some more from the block. This has the most momentum transferred to the block. (This is an elastic collision.) KE and Collisions Totally (Perfectly) Elastic Collision KE is conserved: no sound or heat produced Example: sub atomic particles (electrons), attracting fields http://www.youtube.com/watch?v=x6n-QgjM4Ss (Partially) Elastic Collision some KE is lost as heat or sound Is a broad range containing most collisions Example: billiard balls http://www.youtube.com/watch?v=CgDDiDa3Kzk http://www.youtube.com/watch?v=wFoPawE0LxA KE and Collisions Completely Inelastic Collision KE is conserved: no sound or heat produced Example: sub atomic particles (electrons), repelling fields http://www.youtube.com/watch?v=NN_wwbx6Bew Inelastic Collision more KE is lost as heat or sound the colliding objects stick together Example: Coupling railroad cars http://www.youtube.com/watch?v=qzV8ovAobfE PROBLEM 20 A 2575 kg van runs into the back of a 825 kg compact car at rest. They move off together at 8.5 m/s. Assuming no friction with the ground, find the initial speed of the van. m a 2575kg m b 825kg v ai ? m / s v bi 0 v af 8.5m / s v bf 8.5m / s m a v ai m bv bi m av af m bv bf 2575v ai 825(0) 2575(8.5) 825(8.5) 2575v ai 21887.5 7012.5 2575v ai 28900 v ai 11.2m / s PROBLEM 21 A 5g bullet is fired with a velocity of 100 m/s toward a 10 kg stationary solid block resting on a frictionless surface. What is the change in momentum of the bullet if it becomes embedded in the block? m a .005kg m b 10kg v ai 100m / s v bi 0 v af x v bf x m av ai m bv bi m av af m bv bf .005(100) 10(0) .005( x ) 10( x ) .5 10.005x x .05m / s Change in Momentum of Bullet ma (v af v ai ) .005(.05 100) 0.5kgm / s