1. Copyright Sautter 2003

2. Newton’s Second Law of Motion Acceleration = velocity / time Combining the two equations Rearranging the equation Impulse Momentum

3. Impulse & Momentum As seen on the previous slide, momentum and impulse equations are derived from Newton’s Second Law of Motion (F = ma). Impulse is defined as force times time interval during which the force is applied and momentum is mass times the resulting velocity change. The symbol p is often used to represent momentum, therefore p = mv. The question maybe however, “why develop a momentum equation when it is merely a restatement of F = ma which we already know ?” The answer is that no conservation principles can be applied to forces. There is no such thing as “conservation of force”. However, Conservation of Momentum is a fundamental law which can be applied to a vast array of physics problems. Therefore, manipulating Newton’s Second Law into the impulse – momentum format helps us to implement a basic principle of physics !

4. A collision is a momentum exchange. In all collisions the total momentum before a collision equals the total momentum after the collision. Momentum is merely redistributed among the colliding objects. Momentum = Mass x Velocity p = m x v Σ Momentum before = Σ Momentum after collision collision

5. Momentum & Kinetic Energy Recall the following facts about kinetic energy: Kinetic energy is the energy of motion. In order to possess kinetic energy an object must be moving. As the speed (velocity) of an object increases its kinetic energy increases. The kinetic energy content of a body is also related to its mass. Most massive objects at the same speed contain most kinetic energy. Since the object is in motion, the work content is called kinetic energy and therefore: K.E. = ½ m v 2

6. Types of Collisions (Momentum Transfers) Collisions occur in three basic forms: (1) Elastic collisions – no kinetic energy is lost during the collision. The sum of the kinetic energies of the objects before collision equals the sum of the kinetic energies after collision. It is important to realize that kinetic energy is conserved in these types of collisions. Total energy is conserved in all collisions. (The Law of Conservation of Energy requires it!) The only collisions which are perfectly elastic are those between atoms and molecules. In the macroscopic world some collisions are close to perfectly elastic but when examined closely, they do lose some kinetic energy.

7. Types of Collisions (Momentum Transfers) (2) Inelastic collisions – the objects stick together after collision and remain as a combined unit. Kinetic energy is not conserved ! (3) Partially elastic collisions – kinetic energy is not conserved but the colliding objects do not remain stuck together after collision. Most collisions are of this type. In these collision some of the energy is converted to heat energy or used to deform the colliding objects. That which remains is retained as kinetic energy.

8. Types of Collisions (Momentum Transfers Example collisions: Elastic collisions – pool balls colliding, a golf ball struck with a club, a hammer striking an anvil. Remember, these collisions are not perfectly elastic but they are close! Inelastic collisions – an arrow shot at a pumpkin and remaining embedded, a bullet shot into a piece of wood and not fully penetrating, two railroad cars colliding and coupling together. Partially elastic collisions - two cars colliding and then separating, a softball struck by a bat, a tennis ball hit with a racket.

9. M 1 U 1 + M 2 U 2 = M 1 V 1 + M 2 V 2 CLICK HERE Σ K.E. before collision = Σ K.E. after collision e = 1.0

10. M 1 U 1 + M 2 U 2 = (M 1 + M 2 ) V CLICK HERE Σ K.E. before collision = Σ K.E. after collision / e = 0

11. Measuring Collision Elasticity The elastic nature of a collision can be measured by comparing the K.E. content of the system before and after collision. The closer they are to being equal, the more elastic the collision. Another way to measure elasticity is to use the Coefficient of Restitution. The symbol for this value is e. If e is equal to 1.0 the collision is perfectly elastic. If e is equal to zero the collision is inelastic. If e lies in between 0 and 1.0 the collision is partially elastic. The closer e is to 1.0 the more elastic the collision, the closer e is to 0 the more inelastic.

12. Ball 1 Ball 2 Velocity of ball 1 Before collision u 1 Velocity of ball 2 Before collision u 2 Ball 1 Ball 2 Point of collision Velocity of ball 1 after collision V 1 Velocity of ball 2 after collision V 2

13. Momentum – A Vector Quantity Momentum is a vector quantity (direction counts). In one dimensional collisions motion lies exclusively along the x plane or the y plane. One Dimensional Collisions - The direction of the momentum vectors are identified by the usual conventions of plus on horizontal plane (x axis) is to the right and minus is to the left. In the vertical plane (y axis), plus is up and minus is down. Two Dimensional Collisions – The momentum vectors can be resolved into x – y components and combined using vector addition methods to obtain the momentum sums before and after collision. In two dimensional collisions the sums of the vertical momentum components must be equal before and after collision just as the sums of horizontal momentum components must be equal before and after collision.

14. Σ p before collision = Σ p after collision Σ p x before collision = Σ p x after collision Σ p y before collision = Σ p y after collision

15. Momentum and Explosions In an explosion, momentum is always conserved. Before an explosion occurs, if the exploding object is at rest, the sum of all momentum is zero. P= mv and if v is zero then momentum (p) must be zero ! After the explosion occurs the sum of the momentum must again be zero. This means that the momentum carried out in one direction by a fragment of the exploded item must be balanced by the momentum of another piece carried off in the opposite direction. Since momentum is a vector quantity, this must be true in all directions and in all plane surrounding the exploded object.

16. Since the carts are not in motion before the explosion, the momentum of the system is zero. Therefore after the explosion the momentum of the system must be zero. Σ p before explosion = Σ p after collision The carts are moving in opposite directions after the explosion. The momentum of each of the carts then must be equal but opposite in order to give a sum of zero after the explosion. The more massive cart has a lesser velocity while the smaller cart has a greater velocity so that the momentum of each must be equal ! Play movie Click here To continue

17. 12 1 2 p ’ x1 p ’ x2 p ’ y1 p ’ y2 p x1 p y1 before collision = 0 p x2 and p y2 are both zero (ball 2 is at rest) p = momentum before collision p ’ = momentum after collision p x = p cos θ p y = p sin θ θ1θ1 θ2θ2

18. Momentum Problems A 2400 lb vehicle moving at mph is brought to a stop in 2 seconds. What is the average force acting on the vehicle ? W = mg, m = w/g, m = 2400 lbs/ 32ft/s 2 = 75 slugs V o = 20 mph = (20 x 5280) / 3600 = 29.4 ft/s, V f = 0 Δv = v i – v o = 29.4 – 0 = 29.4 ft / s F x Δt = m x Δv F (2 sec) = 75 slugs x 29.4 ft/s F = 1100 lbs 2400 lbs V o = 20 mphV i = 0 mph

19. Momentum Problems A 40 kg skater traveling at 20 m/s collides with a 60 kg skater moving in the same direction at 2 m/s. (a) what is their velocity if they hold on to each other? (b) how much kinetic energy is lost? (a) Mass # 1 = 40 kg, velocity # 1 = -20 m/s (negative means to the left) Mass # 2 = 60 kg, velocity # 2 = - 2 m/s (negative means to the left) V = velocity of the combined bodies after collision M 1 U 1 + M 2 U 2 = (M 1 + M 2 ) V (40 kg x –20 m/s) + (60 kg x – 2 m/s) = (40 kg + 60 kg) V V = - 9.2 m/s (b) K.E. = ½ mv 2 K.E before collision = ½ (40 kg)(-20 m/s) 2 + ½ (60 kg)(-2 m/s) 2 = 8120 j K.E. after collision = ½ (40kg + 60 kg) (-9.2 m/s) 2 = 4232 j ΔK.E. = 8120 j – 4232 j = 3888 j -20 m/s-2 m/s 40 kg 60 kg V f = ? M 1 U 1 + M 2 U 2 = (M 1 + M 2 ) V

20. Momentum Problems A 40 kg skater traveling at 20 m/s collides with a 60 kg skater moving in the opposite direction at 2 m/s. (a) what is their velocity if they hold on to each other? (b) how much kinetic energy is lost? -20 m/s 40 kg 2 m/s 60 kg M 1 U 1 + M 2 U 2 = (M 1 + M 2 ) V (a) Mass # 1 = 40 kg, velocity # 1 = -20 m/s (negative means to the left) Mass # 2 = 60 kg, velocity # 2 = - 2 m/s (negative means to the left) V = velocity of the combined bodies after collision M 1 U 1 + M 2 U 2 = (M 1 + M 2 ) V (40 kg x –20 m/s) + (60 kg x 2 m/s) = (40 kg + 60 kg) V V = - 6.8 m/s (b) K.E. = ½ mv 2 K.E before collision = ½ (40 kg)(-20 m/s) 2 + ½ (60 kg)(-2 m/s) 2 = 8120 j K.E. after collision = ½ (40kg + 60 kg) (-6.8 m/s) 2 = 2312 j ΔK.E. = 8120 j – 2312 j = 5808 j

21. Momentum Problems A 1000 kg car travels north at 20 m/s and collides with a 1500 kg car moving west at 12 m/s. They stick together. What is the velocity of the wreckage? Resolve the momentum of each car into x and y components P 1x = m 1 v 1 cos 0 0 = 0 P 1y = m 1 v 1 sin 90 0 = 1000kg x 20 m/s x 1.0 = 20,000 kg m /s P 2y = m 2 v 2 sin 180 0 = 0 P 2x = m 2 v 2 cos 180 0 = 1500kg x 12 m/s x 1.0 = 18,000 kg m /s Adding the vectors, P r = ((0 + 18,000) 2 + (20,000 + 0) 2 ) 1/2 P r = 26,900, P r = (m 1 + m 2 ) v, 26,900 kg m/s = (1000kg + 1500 kg) v, v = 10.8 m/s Angle = tan -1 (20,000 / 18,000) = 48 0 north of west 1 20 m/s 2 12 m/s M 1 U 1 + M 2 U 2 = (M 1 + M 2 ) V

22. Momentum Problems A 1.0 kg ball moving at 5 m/s collides head on with a 2.0 kg ball moving at 4 m/s in the opposite direction. The coefficient of restitution is 0.7. Find the velocity of each ball after collision. m 1 = 1 kg, m 2 =2 kg, u 1 = 5 m/s, u 2 = - 4m/s e = (v 2 – v 1 ) / (u 1 – u 2 ), 0.7 = (v 2 – v 1 ) / (5 – (-4)) v 2 – v 1 = 6.3, v 2 = 6.3 + v 1 M 1 U 1 + M 2 U 2 = M 1 V 1 + M 2 V 2, (1 kg x 5 m/s) + (2 kg x (-4 m/s) = (1 kg x v 1 ) + (2 kg) x ( 6.3 + v 1 ) 5 + (-8) = 1v 1 + 12.6 + 2v 1, v 1 = (-15.6) /3 = - 5.2 m/s v 2 = 6.3 + v 1, v 2 = 6.3 + (-5.2) = + 1.1 m/s 1 kg + 5 m/s 2 kg - 4 m/s e = 0.7 M 1 U 1 + M 2 U 2 = M 1 V 1 + M 2 V 2

23. Momentum Problems A 5 kg gun shoots a 15 gram bullet at 600 m/s. What is the recoil velocity of the gun ? Since both the gun and the bullet are at rest before the bullet is fired, the total momentum of the system is zero. The total momentum after firing must also be zero. The mass of the bullet is 15 g = 0.015 kg M b u b + m g u g = m b v b + m g v g (0.015 x 0 ) + (5 x 0) = (0.015 x 600) + 5 v g 5 v g =- 9, v g = - 9 / 5 = - 1.8 m/ s (negative means recoil) + 600 m/s

24. Momentum Problems A ball is dropped from 3 meters and bounces up 1 meter. Find the coefficient of restitution. Object 1 is the ball, object 2 is the ground (which does not move !) e = (v 2 – v 1 ) / (u 1 – u 2 ), u 1 = (2 x 9.8 x 3) 1/2 = -7.7 m/s, u 2 = 0 v 1 = (2 x 9.8 x1) 1/2 = + 4.4 m/s, v 2 = 0 E = (0 – 4.4) / ( - 7.7 – 0) = 0.57 3m 1m Recall: s = (v i 2 – v o 2 ) / 2 a h = (v i 2 – v o 2 ) / 2 g If v o = 0, solve for v i V i = (2gh) 1/2 Down = - Up = +

26. A 60 kg skater pushes off a 50 kg skater at 2.0 m/s. What is the Velocity of the 60 kg skater? (A) 0.6 m/s (B) 1.7 m/s (C) 2.0 m/s (D)2.4 m/s A 1500 kg vehicle moving at 60 km/hr rear ends a 4000 kg vehicle moving at 35 km/hr. They remain attached. What is the velocity of the wreckage ? (A) 9.1 km/hr (B) 42 km/hr (C) 48 km/hr (D) 53 km/hr An 800 kg car going south at 40 km/hr strikes a 1200 kg truck going west at 25 km/hr. What is the speed of the wreckage if the stick together ? (A) 22 km/hr (B) 31 km/hr (C) 33 km/hr (D) 47 km/hr Find the direction the wreckage moves in the previous problem. (A) 20 0 W of S (B) 43 0 W of S (C) 47 0 W of S (D) 70 0 W of S A 5 kg ball moving a 3 m/s collides with a 3 kg ball moving in the same Direction at 2.4 m/s. Find the final speeds of both balls if the e value is 0.8. (A) 2.3 and 2.8 m/s (B) –2.3 and 2.8 m/s (C) – 3 and –2.4 m/s (D) none Click here for answers