Momentum, Energy, Circular Motion Problem Solving Mr. Klapholz Shaker Heights High School

Problem 1 The initial velocity of a baseball (0.20 kg) is 50.0 m s -1. After the ball is hit, its velocity is m s -1. a) What is the change in the velocity of the ball? b) What is the change in the momentum of the ball?

Solution 1 a)  v = v f – v i  v = – 50.0 = m s -1 b) p i = mv i = (0.20)(50.0) = 10 kg m s -1 p f = mv f = (0.20)(-60.0) = -12 kg m s -1  p = p f – p i  p = -12 – 10  p = -22 kg m s -1

Problem 2 (“Explosion”) A skateboarder (40 kg) is holding a bag of potatoes (10 kg) at rest. When the skateboarder throws the potatoes at 8 m s -1, what is the velocity of the skateboarder?

Solution 2 Total Momentum After = Total Momentum Before s: skateboarderp: potatoes p s ’ + p p ’ = p s + p p m s v s ’ + m p v p ’ = m s v s + m p v p (40)v s ’ + (10)v p ’ = (40)v s + (10)v p (40)v s ’ + (10)(8) = (40)(0) + (10)(0) (40)v s ’ + 80 = 0 (40)v s ’ = -80 v s ’ = -2 m s -1

Problem 3 (“Totally Inelastic Collision”) Skateboarder A (40.0 kg) is moving South at 10.0 m/s. Skateboarder B (42 kg) is moving North at 8.0 m/s. When the two skateboarders run into each other, they stick together. a)What is their combined velocity after the impact? b)Was energy conserved?

Solution 3a Total Momentum After = Total Momentum Before p’ = p A + p B (m A + m B )v’ = m A v A + m B v B ( )v’ = (40)(10.0) + (42)(-8.0) 82v’ = 400 – v’ = 64 v’ = m s -1 (South)

Solution 3b Total Energy Before = (½)Mv 2 + (½)Mv 2 = (½)(40)(10) 2 + (½)(42)(8) 2 = = 3344 Joules Total Energy After = (½)(m A +m B )v 2 = (½)(40+42)(0.78) 2 = 24.9 J A lot of heat was made. The mechanical energy is not conserved, but if we included thermal energy, then the total energy would have been conserved.

Problem 4 (“Collision”) Skateboarder A (40.0 kg) is moving South at 10.0 m/s. Skateboarder B (42 kg) is moving North at 8.0 m/s. After the two skateboarders slam into each other, skateboarder A is moving South at 1.0 m/s. What is the velocity of skateboarder B after the impact?

Solution 4 Total Momentum After = Total Momentum Before p A ’ + p B ’ = p A + p B m A v A ’ + m B v B ’ = m A v A + m B v B (40.0)(1.0) + 42v B ’ = (40.0)(10) + (42)(-8.0) v B ’ = 400 – v B ’ = 64 42v B ’ = 24 v B ’ = m s -1 (south)

Problem 5 An egg (0.14 kg) will crack with a force of 0.75 N. If an egg is moving at 11 m/s, how quickly can you stop it without breaking it?

Solution 5 F × T=  p F × T = p 2 – p 1 F × T = mv 2 – mv 1 The least time goes with the greatest force. (-0.75) × T = (0.14)(0) – (0.14)(+11) -0.75T = T = 2.1 s If eggs were more resilient, what would that do to the time?

Problem 6 A ball (0.076 kg) is rolling (4.5 m s -1 ) on a tabletop that is 0.96 m above the floor. Calculate the total energy of the ball.

Solution 6 Total Energy = E k + E p Total Energy = (½)Mv 2 + MgH = (½)(0.076)(4.5) 2 + (0.076)(9.8)(0.96) = = 1.5 J

Problem 7 A ball (0.076 kg) is rolling (4.5 m s -1 ) on a tabletop that is 0.96 m above the floor. The ball rolls off of the edge of the table; how fast is the ball moving when it hits the floor?

Solution 7 Total Energy After = Total Energy Before E k ’ + E p ’ = E k + E p From the previous problem, the total energy before is 1.5 J. E k ’ + E p ’ = 1.5 J (½)Mv’ 2 + MgH’ = 1.5 J (½)(0.076)v’ 2 + (0.076)(9.8)(0) = v’ 2 = 1.5 v’ 2 = v’ = 6.3 m/s (that’s faster than before)

Problem 8 A ball (0.076 kg) is sitting on the floor. How much work would it take to get it rolling at 4.5 m s -1 on a tabletop that is 0.96 m above the floor?

Solution 8 Work = Change in Energy We know that the energy on the floor is 0. The energy on the table is 1.5 J Work = 1.5 – 0 = 1.5 J

Problem 9 How much time does it take a 60.0 W light bulb to do 1.0 Joule of work?

Solution 9 Power = Work ÷ Time 60.0 W = 1.0 J ÷ T T = 1.0 / 60.0 T = s

Problem 10 A ball (50.0 g) is tied to a string and is whirled around in a horizontal circle (radius = 1.2 m) at a constant speed. The ball makes 1.5 revolutions per second. a)Find the acceleration of the ball. b)Find the force that the student must exert on the string.

Solution 10 a) a c = v 2 / R = (?) 2 / R v = dist / time The distance is one and a half circumferences in 1 second. Dist = 1.5×2  R = (1.5)(2)  (1.2) = 11.3 m v = (11.3 m) / (1 s) = 11.3 m s -1 a c = v 2 / R = (11.3) 2 / (1.2) a c = 110 m s -2 (that’s 11 g’s !)

Solution 10 b) F c = ma c F c = (0.050 kg)(110 m/s 2 ) F c = 5.5 N

Problem 11 Flowing Mass problem. A source of water is dropping water onto a balance at the rate of 30 L per minute, from a height of meters. When the water hits the balance the water does not bounce, it just runs off of the pan. What is the reading on the balance?

Solution 11 (1 of 3) F × T=  p F × T = p 2 – p 1 F × T = Mv 2 – Mv 1 ‘v 1 ’ is the speed of the water as it hits the pan. ‘v 2 ’ is the speed after it has been stopped. F × T = M(0) – Mv 1 F × T = -Mv 1 F = -Mv 1 / T F = -{ M / T } × v 1

Solution 11 (2 of 3) F = -{ M / T } × v 1 We can find M/T and v 1. 1 L of water has a mass of 1 kg. So, 30L/min = 30 kg/min = 30kg/60s = 0.5 kg/s M / T = 0.5 kg / s v 2 = u 2 + 2as v 2 = (9.8)(0.5) v = 3.1 m s -1 F = - { M/T }×v 1 = -(0.50kg/s)(3.1m/s) = N

Solution 11 (3 of 3) So if the scale was reading Newtons, then it would read 1.55 N. But most scales read grams. Weight = Mass x g M = W / 9.8 M = 1.55 / 9.8 = kg Mass = 160 grams

Problem 12 How much force is required to produce 100 watts if you are dragging a wagon at 2 m s -1 ?

Solution 12 Power = Force x Speed P = Fv F = P ÷ v F = 100 W / 2 ms -1 F = 50 N

Tonight’s HW: Go through the Mechanics section in your textbook and scrutinize the “Example Questions” and solutions. Bring in your questions to tomorrow’s class.