Download ap fr quiz #16 intro to momentum

AP FR QUIZ #16 INTRO TO MOMENTUM
1: CARS STOPPED BY CONSTANT FORCE BARRIERS—STOPPING TIME
Cars moving along horizontal roads are about to be stopped when they hit a protective barrier. All of the cars are the
same size and shape, but they are moving at different speeds and have different masses. The barriers are all identical
and exert the same constant force.
A
B
10 m/s
m = 3000 kg
20 m/s
C
m = 1000 kg
30 m/s
m = 1000 kg
D
20 m/s
m = 2000 kg
Rank the time that it takes to stop the cars as the barriers apply the same constant force.
Explain your reasoning.
Answer: D > A = C > B. Since the same force is acting on all cars and we are asked about the time interval to bring
the cars to rest, the physical quantity involved is the momentum. The barriers are going to exert an impulse on the
cars bringing them to rest and that impulse will be equal to the change in momentum for the cars. So the ranking
is based on the product of the mass and velocity.
2: BOUNCING CART—CHANGE IN MOMENTUM I
Carts with spring plungers run into fixed barriers. The carts are identical, but are carrying different loads and so have
different masses. The velocity of the cart just before and just after impact is given.
A
C
Before
After
10 kg
10 kg
vo = 4 m/s
vf = 0
Before
After
10 kg
10 kg
vo = 2 m/s
vf = –2 m/s
B
Before
After
10 kg
10 kg
vo = 3 m/s
D
vf = –1 m/s
Before
After
5 kg
5 kg
vo = 5 m/s
vf = –3 m/s
Rank the magnitude of the change in momentum of these carts.
Explain your reasoning.
Answer: All have the same change in momentum, 40 kgm/s to the left. The change in momentum is the final
momentum minus the initial momentum, but in calculating this we have to keep in mind that momentum is a vector
quantity. That means our calculations have to use the signs of the velocity, so for example, the change in
momentum of case B (with positive directions to the right) is –10 (kg)(m)/s –30 (kg)(m)/s = –40 (kg)(m)/s.
3: BOUNCING CART—DIRECTION OF THE CHANGE IN MOMENTUM II
Carts with spring plungers run into fixed barriers. The carts are identical, but are carrying different loads and so have
different masses. The velocity of the cart just before and just after impact is given.
A
C
B
Before
After
10 kg
10 kg
20 kg
vo = 4 m/s
vf = 0
vo = 2 m/s
Before
10 kg
vo = 2 m/s
D
After
Before
vf = 0
Before
After
5 kg
5 kg
10 kg
vf = –2 m/s
After
20 kg
vo = 5 m/s
vf = –3 m/s
(a) Is the direction of the change in momentum in Case A to the left, or to the right? If the change in
momentum cannot be determined, state that explicitly.
Explain.
Answer: Towards the left. The initial momentum is directed right and the final momentum is zero so the change
had to be directed to the left.
(b) Is the direction of the change in momentum in Case B to the left, or to the right? If the change in momentum
cannot be determined, state that explicitly.
Explain.
Answer: Towards the left. The initial momentum is directed right and its final momentum is zero so the change had
to be directed to the left.
(c) Is the direction of the change in momentum in Case C to the left, or to the right? If the change in momentum
cannot be determined, state that explicitly.
Explain.
Answer: Towards the left. The initial momentum is directed right and the final momentum is negative, which here
means to the left, so the change had to be directed to the left.
(d) Is the direction of the change in momentum in Case D to the left, or to the right? If the change in momentum
cannot be determined, state that explicitly.
Explain.
Answer: Towards the left. The initial momentum is directed right and the final momentum is negative, which here
means to the left, so the change had to be directed to the left.
4: BOUNCING CART—MAGNITUDE OF CHANGE IN MOMENTUM III
Carts with spring plungers run into fixed barriers. The carts are identical, but are carrying different loads and so have
different masses. The velocity of each cart just before and just after impact is given.
A
Before
After
10 kg
vo = –3 m/s
C
Before
After
10 kg
20 kg
20 kg
vf = 3 m/s
vo = –2 m/s
Before
After
10 kg
10 kg
vo = –1 m/s
B
D
vf = 1 m/s
vf = 0
Before
After
20 kg
20 kg
vo = –1 m/s
vf = 1 m/s
Rank the magnitude of the change in momentum of the carts.
Explain your reasoning.
Answer: A > B = D > C. The change in momentum is the final momentum minus the initial momentum, but in
doing this we have to keep in mind that momentum is a vector quantity. That means our calculations have to use
the signs of the velocity, so for example, the change in momentum of case B (with positive directions to the right) is
0 (kg)(m)/s –(–40) (kg)(m)/s = 40 (kg)(m)/s. In all four cases we end up with a positive momentum change.
5: FORCE PUSHING BOX—CHANGE IN MOMENTUM
Identical boxes that are filled with different objects are initially at rest. A horizontal force is applied for 10 seconds,
and the boxes move across the floor. The mass of the box with its contents and the net force acting on the box while
the horizontal force is applied is given in each figure.
A
B
F = 75 N
C
F = 50 N
F = 75 N
20 kg
D
10 kg
F = 100 N
10 kg
20 kg
Rank the magnitude of the change in momentum during the 10-second interval for each box.
Explain your reasoning.
Answer: D > A = B > C. The time interval is the same for all four cases, so the magnitudes of the momentum
changes, which are equal to the impulses applied to the boxes, will be proportional to the net forces acting.
6: CARS—CHANGE IN MOMENTUM DURING A CHANGE OF VELOCITY
Before and after "snapshots" of a car's velocity are shown. All cars have the same mass.
A
B
Before
After
+10 m/s
0
C
Before
After
+10 m/s
–10 m/s
D
Before
After
Before
After
+20 m/s
+20 m/s
–10 m/s
–20 m/s
Rank the magnitude of the change in momentum of the cars.
Explain your reasoning.
Answer: B > A = D > C. Since the masses are all the same the magnitudes of the momenta changes will be equal to
the difference in the velocities, keeping in mind that velocity is a vector quantity. For example, in A the change in
velocity is (0 – 10)m/s = -10 m/s and in D it is (-20 - -10)m/s = -10 m/s.
7: OBJECT CHANGING VELOCITY I—IMPULSE
A 2-kg object accelerates as a net external force is applied to it. During the 5-second interval that the force is applied,
the object’s velocity changes from 3 m/s to the right to 7 m/s to the left.
3 m/s
5 seconds later:
7 m/s
A student states:
“The change in momentum of this object during these 5 seconds was 8 kg·m/s so the impulse applied to
this object during these 5 seconds was 8/5 kg·m/s.”
What, if anything, is wrong with this statement? If something is wrong, identify and explain how to correct all
errors. If this statement is correct, explain why.
Answer: This statement is incorrect because (1) momentum is a vector so a direction should be specified for the
change in momentum, (2) impulse = change in momentum so the time does not enter into the problem, and (3) the
actual change in momentum (taking right as positive) is (2kg)(–7m/s) – (2kg)(+3m/s) = –20 kg•m/s, or 20 kg•m/s to
the left.
8: OBJECT CHANGING VELOCITY II—IMPULSE
A 2-kg object accelerates as a net external force is applied to it. During the 5-second interval that the force is applied,
the object’s velocity changes from 3 m/s to the right to 7 m/s to the left.
3 m/s
5 seconds later:
7 m/s
A student states:
“The change in velocity for this 2 kg object was 4 m/s, so the change in momentum, and also the impulse,
was 8 kg·m/s.”
What, if anything, is wrong with this statement? If something is wrong, identify and explain how to correct it.
If this statement is correct, explain why.
Answer: This statement is incorrect because momentum is a vector so the change in momentum (taking to the right
as the positive direction) is equal to 2 kg (7 m/s  (+3 m/s)) = 20 kgm/s, or 20 kgm/s to the left.
9: OBJECT CHANGING VELOCITY III—IMPULSE
A 2-kg object accelerates as a net external force is applied to it. During the 5-second interval that the force is applied,
the object’s velocity changes from 3 m/s to the right to 7 m/s to the left.
3 m/s
5 seconds later:
7 m/s
Several students discussing the impulse on this object state the following:
Andre:
Bela:
“The impulse is equal to the change in momentum, which is (2 kg)(3 m/s + 7 m/s) = 20 kg·m/s.”
“But the change in velocity is 4 m/s. We multiply by the mass to get the change in momentum, and also the
impulse, which is 8 kg·m/s.”
Carleton: “The change in momentum of this object during these 5 seconds was 8 kg·m/s so the impulse applied to
this object during these 5 seconds was 8/5 kg·m/s.”
Dylan: “The impulse is the force F times the time t and since we don’t know the force, we can’t find the impulse
for this situation.”
With which, if any, of these students do you agree?
Andre _____ Bela _____ Carleton _____ Dylan _____ None of them______
Explain your reasoning.
Answer: None of these students are correct. The change in velocity is the final velocity minus the initial velocity, or
10 m/s leftward. So the change in momentum is 20 kgm/s left. Andre has the correct magnitude but has forgotten
to include the direction of this vector quantity and his calculation does not use the correct velocities.
10: OBJECT CHANGING DIRECTION AND SPEED—IMPULSE
A student proposes the following description for the impulse on a 2-kg object that changes direction and speed as
shown:
“The object goes from moving at 3 m/s in the positive x-direction to 7 m/s in the positive y-direction in 5
seconds. So the impulse given to it is 8 kg·m/s, since the impulse equals the change in momentum. The 5
seconds does not enter into the calculation of this impulse.”
3 m/s
5 seconds later:
7 m/s
What, if anything, is wrong with this statement? If anything is wrong, identify it and explain how to correct it.
If this statement is correct, explain why.
Answer: The 8 kgm/s is incorrect since the impulse and the change in momentum are vector quantities and this
description does not treat them as such. Using a coordinate axis with positive x to the right and positive y toward
the top of the page, the final momentum of the object is zero in the x-direction, so the change in momentum in the
x-direction is –6 kgm/s. The initial momentum of the object in the y-direction is zero, so the change in momentum
in the y-direction is 14 kgm/s. We can use the Pythagorean theorem to find the magnitude of the change in
momentum, which is 15.2 km/s. The direction of the change in momentum is 23.2 degrees to the left of the
positive-y axis.
11: FORCE VS. TIME GRAPH—IMPULSE APPLIED TO BOX
A 10-kg box, initially at rest, moves along a frictionless horizontal surface. A horizontal force to the right is applied to
the box. The magnitude of the force changes as a function of time as shown.
Force (Newtons)
3
2
1
Time (seconds)
2
4
6
8
10
A student calculates that the impulse applied by the force during the first 2 seconds is 4 N·s and that the impulse
applied during the following 3 seconds is 6 N·s.
What, if anything, is wrong with these calculations? If something is wrong, identify it and explain how to
correct it. If these calculations are correct, explain why.
Answer:
The impulse is given by the area under the curve which for the first two seconds is 2 N·s since (1/2)·2 s·2 N= 2 N·s.
The student’s calculation for the next three seconds is correct.
12: FORCE VS. TIME GRAPH—IMPULSE APPLIED TO BOX
A 10-kg box, initially at rest, moves along a frictionless horizontal surface. A horizontal force to the right is applied to
the box. The magnitude of the force changes as a function of time as shown.
Force (Newtons)
3
2
1
Time (seconds)
2
4
6
8
10
Rank the impulse applied to the box by this force during each 2-second interval indicated below.
A. 0 to 2 s
B. 2 to 4 s
C. 4 to 6 s
D. 6 to 8 s
E. 8 to 10 s
OR
1
Greatest
2
3
4
5
Least
All
the same
All
zero
Cannot
determine
Explain your reasoning.
Answer: B > C > A > D > E.
The impulse for each time interval is equal to the area under the graph during that time interval. From zero to 2
seconds this is 2 kg m/s; from 2 to 4 seconds it is 4 kg m/s; from 4 to
6 seconds it is 3.67 kg m/s; from 6 to 8 seconds it is 1.33 kg m/s; and from 8 to 10 seconds it is zero. (Note that you
don’t really need to calculate values as a visual inspection will enable one to rank the areas.)