AP Physics 1 - Review 03 - Rubric Scarborough High School SCIENCE 1

AP Physics 1: Review Packet 03 RUBRIC
Problem 1 Problem 2
(a) (a) (c) Both trains have the same net force
 Fw (or Fg or mg) is straight down  The horse pulls the cart forward  Both trains have the same net force
 FN is straight up  The cart moves forward  Net force = mass  acceleration
 Ff is straight to the left  The cart pulls backward on the horse  Acceleration is the slope
 Ff = ma  The horse doesn’t go forward as  Accel. is the slope of velocity vs. time
 FN = mg quickly as it would without the cart  Both graphs have the same slope
 FN = mg > Ff  Both graphs have the same accel.
(b)  Slope is constant so acceleration is
(b)  The heavy car pushes the light car constant
 Fw (or Fg or mg) is straight down forward
 FN is up-right perpendicular to plane  The light car moves foward (d) Weight > Norm. and Tens. > Frict.
 Ff is up-left parallel to plane  The light car pushes the heavy car  Weight is greater than normal
 Ff = mgsin backward  Tension is greater than friction
 FN = mgcos  The heavy car slows down  There is one down force (weight)
 mg > FN > Ff  There are two up forces (normal and
(c) vertical component of tension)
 The helicopter pushes down on the air  Weight must be greater than normal
(c)
 Fw (or Fg or mg) is straight down  The air flows downward since weight balances two forces
 FN is up-right perpendicular to plane  The air pushes up on the helicopter  Horizontal component of tension
 Ff is down-right parallel to plane  The helicopter remains above the equals friction
 Fa is up-left parallel to plane ground  Tension has a vertical component as
 Fa = Ff + mgsin well
Problem 3  Tension has two components making
 FN = mgcos (a) Spring stretches the same it bigger than friction
 Weight is ranked highest  Stretch of the spring is same in both
 Friction is ranked lowest cases (e) Student C is correct
 To change the stretch of the spring,  Student C is correct
(d) the spring must apply a different force  Student A is wrong: Forces can’t have
 Fw (or Fg or mg) is straight down  To apply a different force, the block the same length because the system is
 FN is straight to the left (and/or elevator) must be accelerating. accelerating
 Ff is straight up  “Constant speed” means zero  System is accelerating means forces
 Fa is straight to the right acceleration. can’t balance
 Fa = FN  Zero acceleration in both cases, so the  Student B is wrong: 2 kg slows down
 Ff = mg spring exerts the same force in both  Force opposing velocity is greater
 Weight equals friction cases. than force with velocity if an object
 Normal equals applied  Therefore the stretch of the spring is slows down
 FN and Fa greater than Fw and Ff the same.  Student C is correct: 5 kg moves right
so friction must be to the left.
(e) (b) Linda would get pulled up off of the  Friction opposes the direction the
 Fw (or Fg or mg) is straight down ground block slides.
 FN is straight up  Linda would get pulled up off of the
 mg – FN = mv2/r ground
 mg > FN  The barrel is heavier than she is
 The Linda-Barrel system would have
(f) a net force causing the barrel to go
 Fw (or Fg or GmM/r2) is straight left down and Linda to go up
 GmM/r2 = mv2/r




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