Goal: Reasoning with forces and kinematics
Source: UMPERG-ctqpe60 variant
Two
blocks, M2 = M1 but of different sizes, having the
same speed, move from a frictionless surface onto a surface having
friction coefficient μk.
Which stops in the shorter time?
Goal: Problem solving
Source: UMPERG-ctqpe44-46
A
block slides along a frictionless surface and onto a slab with a rough
surface. The plot on the right shows the velocity of the blue slab as a
function of time. The slab has mass of 4kg and the block has mass of
2kg. What is the friction force on the small block at t = 1 second?
(5) The acceleration of the slab can be found from the plot. The
only force on the slab in the horizontal direction is the friction force
so it must be responsible for the acceleration. The force on the block
can then be found using the 3rd law.
Goal: Reasoning with dynamics
Source: UMPERG-ctqpe43
Block
m1 sits on block m2 and both sit on the floor of
an elevator at rest. When the elevator starts to move down, the normal
force on the upper block will …
(3) As it starts the elevator must accelerate downward and so
will the upper block. The only forces on the block are gravity and the
normal force. The normal force must diminish so gravity can provide the
downward acceleration.
Students answering #2 may have interpreted the question to mean ‘as the
elevator moves’ and think that the elevator moves with constant velocity.
Goal: Reasoning with dynamics
Source: UMPERG-ctqpe30
A
block of mass m, when placed on a rough inclined plane and moved, moves
down the plane with constant speed. If a block of mass 2m were placed
on the same incline and moved, it would …
The block will have the same motion. Both the gravitational force
and the friction force scale with the mass so there is no net force in
either case.
Goal: Hone the concept of static friction
Source: UMPERG-ctqpe27
A mass of 5 kg sits at rest on an incline making an angle of 30° to
the horizontal.
If μs = 0.7 the friction force on the block is
(2) this is all that is needed to hold the block at rest. Some
students will give #4 as the answer having calculated the maximum static
friction force.
It helps to classify forces as model forces obtainable from a formula,
and procedure forces. Static friction is an example of a procedure
force, one that cannot be determined without application of the 2nd law.
Goal: Problem solving with dynamics
Source: UMPERG-ctqpe21
Two blocks rest on a frictionless surface. Both blocks move to the right
with acceleration of 2 m/s2. The force on the big block due
to the small block is
(7) The force on the small block must cause the specified acceleration.
The 3rd law requires that the force on the big block be equal and
opposite. The magnitude is 4N but it is directed to the left.
Goal: Reasoning with dynamics.
Source: UMPERG
A child stands on a spinning disk. Suppose that there is friction
between the child’s shoes and the surface of the disk. While holding a
rock the child stands at the largest radius possible for the current
angular velocity without slipping. After releasing the rock, the child
will…
(4) There should be no consequence of dropping the rock. Because the
normal force changes, so does the friction force. The new friction force
is still able to provide the necessary centripetal force for the
circular motion.
Goal: Problem solving
Source: UMPERG
A bug sits on a disk at a point 0.5 m from the center. If the
coefficient of friction between the bug and disk is 0.8, the maximum
angular velocity the disk can have before the bug slips off the disk is
most nearly:
(2) Some students may respond (6) thinking that the mass of the bug is
needed for solution.
Goal: Hone the concept of impulse and recognize an application of the 3rd law.
Source: UMPERG-ctqpe92
Compare two collisions that are perfectly inelastic. In case (A) a car
traveling with velocity V collides head-on with a sports car having half
the mass and traveling in the opposite direction with twice the speed.
In case (B) a car traveling with velocity V collides head-on with a
light truck having twice the mass and traveling in the opposite
direction with half the speed. In which case is the impulse delivered
to the car during the collision the greatest?
(3) The impulse delivered to the automobile is the same in both cases.
In both cases the initial momentum of the automobile is MV to the right
and the final momentum is zero.
Impulse is related to the change in momentum. This question provided the
opportunity to discuss the definition of impulse [integral of force over
time interval] and its relation to momentum change. Many students think
I=Δp is the definition of impulse rather than the result of Newton’s
second law. Students should realize that no statement can be made about
the forces exerted on the two cars – only that the integral of the force
over the collision time is the same.
How do the forces acting on the car in the two cases compare? Which
collision takes longer?
Set up the comparison with collision carts.
Goal: Reasoning
Source: UMPERG
Consider the arrangement of pulleys and masses shown below. The masses
of the pulleys are small. Ignore friction.
This system is initially at rest. What will happen if the ring R is
moved to the right?
(3) If the ring is moved to the right the upward force on M is
decreased,so M will accelerate downward. Initially the tension is Mg/2.
When the strings are at an angle the tension is insufficient to support
M.
Answers are not as important as approach. What did students do to
understand the physical situation? Did they draw pictures? Did they draw
a free-body diagram?
Does the tension stay the same? …increase? …decrease? After moving
the ring, would I need a smaller or larger mass M to keep the system
from moving?
After students make predictions and discuss their reasoning have
students vote a second time. Then demonstrate what happens.
Commentary:
Answer
(3) Both blocks will experience the same acceleration. If they
have the same initial velocity, they will stop in the same time.