Download Unit 1: Kinematics

Physics 4 Review
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Name _________________________
Understand the concept of work and power.
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Work, W = F||d, measured in joules (W = Fcosd)
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When F  and d , then + work
When F  and d , then – work
When F  and d , then 0 work (planet orbit)
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Work is a scalar (+ or – refers to amount)
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Work equals area under the F vs. d graph
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Power, P = W/t = Fvav, measured in watts
Understand the concept of mechanical energy.
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Kinetic energy, K = ½mv2, measured in joules
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Gravitational potential energy, Ug = mgh, when close
to the earth's surface or Ug = -GMm/r between planets
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Spring potential energy, Us = ½kx2
Contrast conservative and non-conservative forces.
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Conservative forces, such as gravity and spring
convert potential energy  kinetic energy
o W = F||d, where d = displacement
o When F  and d , then U  K
When F  and d , then K  U
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Non-conservative forces, such as friction and
push/pull change the total mechanical energy
o W = F||d, where d = distance
o When F  and d , then increase Energy (+W)
When F  and d , then decrease Energy (-W)
Work/Mechanical Energy model
Spring System
Us = ½kx2
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Spring Force, Fs
(Conservative force)
Object
K = ½mv2
Gravitational Force, Fg
(Conservative force)
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Gravity System
Ug = mgh
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Push or Pull Force, Fp
(Non-conservative force)
Earth
Solving conservation of mechanical energy problems.
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Problems with conservative forces only
½mv12 + mgh1 + ½kx12 = ½mv22 + mgh2 + ½kx22
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problems with non-conservative forces
K1 + U1 ± Wp – Wf = K2 + U2
½mv12 + mgh1 + ½kx12 ± F||d = ½mv22 + mgh2 + ½kx22
Understand linear momentum and its relationship to force
and kinetic energy.
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Linear momentum, p = mv, measured in kg•m/s
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Momentum is a vector (+ or – refers to direction)
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Impulse force, J = Ft = mv = p
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kinetic energy  momentum is K = p2/2m
Solve conservation of linear momentum problems.
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Momentum is conserved when there are no external
forces involved
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Explosion problems: (mA + mB)v = mAvA' + mBvB'
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Inelastic collision, where objects stick after colliding:
mAvA + mBvB = (mA + mB)v'
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Elastic collision (two formulas): vA + vA' = vB + vB' and
mAvA + mBvB = mAvA' + mBvB'
Solve collisions in two dimensions problems.
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mAvAx + mBvBx = (mA + mB)vx' or mAvAx' + mBvBx'
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mAvAy + mBvBy = (mA + mB)vy' or mAvAy' + mBvBy'
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Solve an energy problem with conservation forces only.
A spring (k = 1000 N/m) is compressed 0.10 m and is used
to launch a 2.0-kg ball upward from a table top 1.0-m
above the floor. The spring takes 0.15 s to launch the ball.
a. What is the ball's velocity just after it is launched?
b. How much power is generated by the spring?
c. How high does the ball go?
d. How fast is the ball traveling when it hits the floor?
e. How fast would the ball be traveling when it hits the
floor if it was initially launched horizontally?
Solve an energy problem with non-conservative forces.
The same spring (k = 1000 N/m) is used to launch a 5.0-kg
block. The spring is compressed 0.20 m and released. The
block first slides along a frictionless surface then across a
rough section of flooring ( = 0.30).
a. Graph Fs vs. distance for the decompressing spring.
(1) What does the slope of the graph represent?
(2) What does the area under the graph represent?
b. How much work is done on the block by the spring
(include sign in the answer)?
c. What is the velocity of the block just after it is released
from the spring?
d. What is the force of friction on the block?
e. How far does the block slide along the rough flooring
before it comes to a halt?
Solve an Inelastic collision problem.
A 3-kg block slides on a frictionless table at 5 m/s and
strikes a 5-kg block at rest on the edge of a table. The two
blocks stick together and slide off of the table and strike
the floor 1 m below.
a. What is the total momentum of the two blocks before
the collision?
b. What is the velocity of the combined blocks after the
collision?
c. How much time are the two blocks in the air?
d. How far from the base of the table do the two blocks
land?
Solve an semi-elastic collision problem.
A 1-kg puck traveling due east at 3 m/s strikes a 3-kg pluck
at rest. The 1-kg block slides at 2 m/s in the direction of
30o south of east.
a. Complete the table.
puck
before the collision
after the collision
1 kg
3 kg
1 kg
3 kg
px
py
b. What is the resulting velocity of the 3-kg puck after the
collision?
c. What is the resulting direction of the 3-kg puck after
the collision?
d. What is the percentage of the original kinetic energy is
conserved after the collision?
Solve ballistics problems.
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bullet strikes a heavy block, which either swings to a
maximum height or slides a distance along a surface
inelastic collision between bullet (m) and block (M):
mvbullet = (M + m)v'
Conservation of energy, swing:
½(M + m)v'2 = (M + m)gh
Conservation of energy, slide:
½(M + m)v'2 = (M + m)gd