momentum, energy, circular motion
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Mr. Klapholz Shaker Heights High School. Momentum, Energy, Circular Motion. Problem Solving. 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 -60.0 m s -1 . a) What is the change in the velocity of the ball? - PowerPoint PPT PresentationTRANSCRIPT
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Momentum, Energy,Circular Motion
Problem Solving
Mr. KlapholzShaker Heights
High School
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Problem 1The initial velocity of a baseball (0.20 kg) is 50.0 m s-1. After the ball is hit, its velocity is -60.0 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?
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Solution 1a)
Dv = vf – vi
Dv = -60.0 – 50.0= -110. m s-1
b) pi = mvi = (0.20)(50.0) = 10 kg m s-1
pf = mvf = (0.20)(-60.0) = -12 kg m s-1 Dp = pf – pi
Dp = -12 – 10Dp = -22 kg m s-1
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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?
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Solution 2Total Momentum After = Total Momentum Before
s: skateboarder p: potatoesps’ + pp’ = ps + pp
msvs’ + mpvp’ = msvs + mpvp
(40)vs’ + (10)vp’ = (40)vs + (10)vp
(40)vs’ + (10)(8) = (40)(0) + (10)(0)(40)vs’ + 80 = 0
(40)vs’ = -80vs’ = -2 m s-1
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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?
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Solution 3aTotal Momentum After = Total Momentum Before
p’ = pA + pB
(mA + mB)v’ = mAvA + mBvB
(40.0 + 42)v’ = (40)(10.0) + (42)(-8.0)82v’ = 400 – 336
82v’ = 64v’ = +0.78 m s-1 (South)
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Solution 3bTotal Energy Before = (½)Mv2 + (½)Mv2
= (½)(40)(10)2 + (½)(42)(8)2 = 2000 + 1344 = 3344 Joules
Total Energy After = (½)(mA+mB)v2 = (½)(40+42)(0.78)2
= 24.9 JA 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.
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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?
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Solution 4Total Momentum After = Total Momentum Before
pA’ + pB’ = pA + pB
mAvA’ + mBvB’ = mAvA + mBvB
(40.0)(1.0) + 42vB’ = (40.0)(10) + (42)(-8.0) 40.0 + 42vB’ = 400 – 336
40 + 42vB’ = 6442vB’ = 24
vB’ = +0.57 m s-1 (south)
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Problem 5An 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?
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Solution 5F × T= Dp
F × T = p2 – p1 F × T = mv2 – mv1
The least time goes with the greatest force.(-0.75) × T = (0.14)(0) – (0.14)(+11)
-0.75T = -1.54 T = 2.1 s
If eggs were more resilient, what would that do to the time?
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Problem 6A 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.
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Solution 6Total Energy = Ek + Ep
Total Energy = (½)Mv2 + MgH= (½)(0.076)(4.5)2 + (0.076)(9.8)(0.96)
= 0.77 + 0.71= 1.5 J
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Problem 7A 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?
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Solution 7Total Energy After = Total Energy Before
Ek’ + Ep’ = Ek + Ep
From the previous problem, the total energy before is 1.5 J.
Ek’ + Ep’ = 1.5 J (½)Mv’2 + MgH’ = 1.5 J
(½)(0.076)v’2 + (0.076)(9.8)(0) = 1.5 0.038v’2 = 1.5
v’2 = 39.46 v’ = 6.3 m/s (that’s faster than before)
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Problem 8A 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?
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Solution 8Work = 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
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Problem 9How much time does it take a 60.0 W light bulb to do 1.0 Joule of work?
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Solution 9Power = Work ÷ Time
60.0 W = 1.0 J ÷ TT = 1.0 / 60.0
T = 0.017 s
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Problem 10A 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.
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Solution 10a)
ac = v2 / R = (?)2 / Rv = dist / time
The distance is one and a half circumferences in 1 second.
Dist = 1.5×2pR = (1.5)(2)p(1.2) = 11.3 mv = (11.3 m) / (1 s) = 11.3 m s-1
ac = v2 / R = (11.3)2 / (1.2)ac = v2 / R = (11.3)2 / (1.2)
ac = 110 m s-2 (that’s 11 g’s !)
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Solution 10b)
Fc = mac Fc = (0.050 kg)(110 m/s2)
Fc = 5.5 N
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Problem 11Flowing Mass problem.A source of water is dropping water onto a balance at the rate of 30 L per minute, from a height of 0. 50 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?
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Solution 11 (1 of 3)F × T= Dp
F × T = p2 – p1 F × T = Mv2 – Mv1
‘v1’ is the speed of the water as it hits the pan.‘v2’ is the speed after it has been stopped.
F × T = M(0) – Mv1 F × T = -Mv1 F = -Mv1 / T
F = -{ M / T } × v1
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Solution 11 (2 of 3)F = -{ M / T } × v1
We can find M/T and v1.1 L of water has a mass of 1 kg.
So, 30L/min = 30 kg/min = 30kg/60s = 0.5 kg/sM / T = 0.5 kg / s
v2 = u2 + 2asv2 = 02 + 2(9.8)(0.5)
v = 3.1 m s-1
F = - { M/T }×v1 = -(0.50kg/s)(3.1m/s) = -1.55 N
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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 gM = W / 9.8
M = 1.55 / 9.8 = 0.158 kgMass = 160 grams
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Problem 12How much force is required to produce 100 watts if you are dragging a wagon at 2 m s-1?
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Solution 12Power = Force x Speed
P = F•vF = P ÷ v
F = 100 W / 2 ms-1 F = 50 N
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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.