objectives 1. Identify different types of collisions. 2. Determine how much kinetic energy is lost in perfectly inelastic collisions. 3. Compare conservation of momentum and conservation of kinetic energy in perfectly inelastic and elastic collisions. 4. Find the final velocity of an object in perfectly inelastic and elastic collisions.

Types of collisions Inelastic collision elastic collision

Elastic and inelastic collisions An inelastic collision is a collision in which the kinetic energy of the system of objects is not conserved. It is transformed into other non-mechanical forms of energy such as heat energy and sound energy. In an elastic collision, the two objects bounce off each other, and kinetic energy is conserved.

A special case of inelastic collision – perfectly inelastic collision The two objects stick together and have the same speed after collision m1v1 + m2v2 = m1v1’ + m2v2’ Since the two objects stick together after the collision, v1’ = v2’ = v m1v1 + m2v2 = (m1+ m2)v’

Example 1 Cart A (50. kg) approaches cart B (100. kg initially at rest) with an initial velocity of 30. m/s. After the collision, cart A locks together with cart B. both travels with what velocity? Before collision after collision B A B A mA = 50 kg, vA = 30. m/s mB = 100. kg, vB = 0 vA’ = vB’ = v’ = ? mAvA + mBvB = mAvA’ + mBvB’ (50.kg)(30.m/s) + (100.kg)(0) = (50.kg +100.kg)v’ v’ = 10. m/s mass increases by 3 times (50 kg to 150 kg), speed decrease by 3 times (30 m/s to 10 m/s)

Example 2 A railroad diesel engine coasting at 5.0 km/h runs into a stationary flatcar. The diesel’s mass is 8,000. kg and the flatcar’s mass is 2,000. kg. Assuming the cars couple together, how fast are they moving after the collision? Before collision after collision Flat car diesel Flat car diesel v’ = 4.0 km/h mass increases by 1.25 times (8000 kg to 10000 kg), speed decrease by 1.25 times (50 m/s to 40 m/s)

Example 3 (pay attention to directions) Cart A (50. kg) moving with an initial velocity of 30. m/s approaches cart B (100. kg moving with initial velocity of 20. m/s towards cart A. The two carts lock together and move as one. Calculate the magnitude and the direction of the final velocity. Before collision after collision B A B A v’ = 3.3 m/s to the left

Example 4 A block of mass M initially at rest on a frictionless horizontal surface is struck by a bullet of mass m moving with horizontal velocity v. What is the velocity of the bullet-block system after the bullet embeds itself in the block? Before collision after collision mv + M(0) (m + M)v’ mv + 0 = (m + M)v’ mv / (M+m) = v’

Example 5 A woman with horizontal velocity v1 jumps off a dock into a stationary boat. After landing in the boat, the woman and the boat move with velocity v2. Compared to velocity v1, velocity v2 has the same magnitude and the same direction the same magnitude and opposite direction smaller magnitude and the same direction larger magnitude and the same direction

Example 6 – Sample Problem 6E A 1850 kg luxury sedan stopped at a traffic light is struck from the rear by a compact car with a mass of 975 kg. the two cars become entangle as a result of the collision. If the compact car was moving at a velocity of 220 m/s to the north before the collision, what is the velocity of the entangled mass after the collision? 7.59 m/s North

Class work Page 224 6E #1-5 Pp 234-235 #31-36, 44*, 45, 48 3.8 m/s S 4.27 m/s N 4.2 m/s Right a. 3.0 kg b. 5.32 m/s Pp 234-235 #31-36, 44*, 45, 48 1 m/s 3.00 m/s 4.2 m/s a. 1.80 m/s b. 2.16 x 104 J a. 0.81 m/s E; b. 1400 J 6.00 kg 23 m/s 14.5 m/s N

Example 7 - Kinetic energy is not constant in inelastic collisions Two clay balls collide head-on in a perfectly inelastic collision. The first ball has a mass of 5. 00 kg and an initial velocity of 4.00 m/s to the right. The mass of the second ball is 0.250 kg., and it has an initial velocity of 3.00 m/s to the left. What is the final velocity of the composite ball of clay after the collision? What is the decrease in kinetic energy during the collision?

Class work Page 226 - 6F #1-3 Pp 234-235 #34-35, 49 a. 0.43 m/s W b. 17 J a. 6.2 m/s S b. 3 J a. 4.6 m/s S b. 3.9 x 103 J Pp 234-235 #34-35, 49 a. 1.80 m/s b. 2.16 x 104 J a. 0.81 m/s E; b. 1400 J a. 2.1 m/s E; b. 4.1 x 104 J

Elastic Collision In an elastic collision, two objects collide and return to their original shapes with no change in total kinetic energy. Most collisions are neither elastic nor perfectly inelastic. Elastic and perfectly inelastic collisions are limiting cases. However, in this book, we assume that when objects bounces off each other, the collision is elastic. Momentum and Kinetic energy remain constant elastic collisions.

Example 8 A 0.015 kg marble moving to the right at 0.225 m/s makes an elastic head-on collision with a 0.930 kg shooter marble moving to the left at 0.180 m/s. after the collision, the smaller marble moves to the left at 0.315 m/s. Assume that neither marble rotates before or after the collision and that both marbles are moving on a frictionless surface. What is the velocity of the 0.030 kg marble after the collision? Confirm your answer by calculating the total kinetic energy before and after the collision.

Another special case of collision - Explosions Total system momentum is conserved for collisions between objects in an isolated system, there are no exceptions to this law. This same principle of momentum conservation can be applied to explosions. Momentum before the explosion is zero. so the momentum after the explosion is also zero. m1v1 + m2v2 = m1v1’ + m2v2’ 0 = m1v1’ + m2v2’

Consider a homemade cannon. p(after) = 0 p(cannon) + p(ball) = 0 p(before) = 0 In the exploding cannon, total system momentum is conserved. The system consists of two objects - a cannon and a tennis ball. Before the explosion, the total momentum of the system is zero since the cannon and the tennis ball located inside of it are both at rest. After the explosion, the total momentum of the system must still be zero. If the ball acquires 50 units of forward momentum, then the cannon acquires 50 units of backwards momentum. The vector sum of the individual momentum of the two objects is 0. Total system momentum is conserved.

Example 8 A 2.0-kilogram toy cannon is at rest on a frictionless surface. A remote triggering device causes a 0.005-kilogram projectile to be fired from the cannon. Which equation describes this system after the cannon is fired? mass of cannon + mass of projectile = 0 speed of cannon + speed of projectile = 0 momentum of cannon + momentum of projectile = 0 velocity of cannon + velocity of projectile = 0

Example 9 A 2-kilogram rifle initially at rest fires a 0.002-kilogram bullet. As the bullet leaves the rifle with a velocity of 500 meters per second, what is the momentum of the rifle-bullet system?

Example 10 A 56.2-gram tennis ball is loaded into a 1.27-kg homemade cannon. The cannon is at rest when it is ignited. Immediately after the impulse of the explosion, a photogate timer measures the cannon to recoil backwards a distance of 6.1 cm in 0.0218 seconds. Determine the post-explosion speed of the cannon and of the tennis ball. Given: Cannon: m = 1.27 kg; d = 6.1 cm; t = 0.0218 s Ball: m = 56.2 g = 0.0562 kg The strategy for solving for the speed of the cannon is to recognize that the cannon travels 6.1 cm at a constant speed in the 0.0218 seconds.   vcannon = d / t = (6.1 cm) / (0.0218 s) = 280 cm/s (rounded The strategy for solving for the post-explosion speed of the tennis ball involves using momentum conservation principles. mball • vball = - mcannon • vcannon vball = 63.3 m/s

Example 11 A 60. kg man standing on a stationary 40. kg boat throws a .20 kg baseball with a velocity of 50. m/s. With what speed does the boat move after the man throws the ball? Assume no friction between the water and the boat. v’(boat) = -0.1 m/s (in the opposite direction of the baseball)

Example 12 In the diagram, a 100.-kilogram clown is fired from a 500.-kilogram cannon.  If the clown's speed is 15 meters per second after the firing, then what is the recoil speed (v) of the cannon? 3.0 m/s