Physics
Momentum “Bucket” Model
I. Overview
This document provides an overview of the “Momentum Bucket” problem
solving strategy. This can be introduced during Chapter 1, Activity 8
“Concentrating on Collisions” to aid students in solving problems involving the
Law of Conservation of Momentum. These types of problems usually involve a
collision between two or more objects either of which may be moving before
and/or after the collision.
This guide will introduce the “Momentum Bucket Model” problem solving
strategy in Part II, provide an example of this problem solving strategy in Part III,
and provide a student sample response in Part IV. A list of supplemental
problems is included in Part V of the document.
II. Momentum “Bucket” Model Problem Solving Strategy
A problem solving strategy for addressing problems involving the Law of
Conservation of Momentum is given below.
Step 1: Statement of Conservation of Momentum
This step requires the student to state the Law of Conservation of Momentum:
pi = pf. Although a small step, it requires the student to state which model he will
be using.
Step 2: Define the positive direction (the organizational part)
Since momentum is a vector quantity, it is important to define the positive
direction for motion, especially when the two objects in motion are moving in
opposite directions. The direction for the positive direction is arbitrary, but in this
document the positive direction will be defined to be towards “right” or the
direction consistent with the positive x-axis as defined in Algebra I classes.
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Step 3: Draw Momentum Buckets
The concept of Conservation of Momentum requires the student to analyze the
INSTANT BEFORE the collision and the INSTANT AFTER the collision. Here students
fill each “momentum bucket” depending on whether or not each object is
moving (i.e. possesses momentum). There are three possible cases. Each is
described below:
Case I: One or both objects are traveling in the positive direction BEFORE the
collision; one or both objects is traveling in the positive direction AFTER the
collision. In the case shown below, both m1 and m2 are traveling to the right
before the collision, only m2 is moving after the collision.
After collision
Before collision
v1i
m1
v2i
m2
+
pm1
m1
=
pm2
m2
+
pm2
in
The “i” and “f” subscripts represent the “initial” and “final”, representing the
instant before and after the collision, respectively.
m1
v1i
v1f
=
=
=
the mass of object 1 [kg]
the velocity of object 1 at the instant BEFORE the collision [m/s]
the velocity of object 1 at the instant AFTER the collision [m/s]
m2
v2i
v2f
=
=
=
the mass of object 2 [kg]
the velocity of object 2 at the instant BEFORE the collision [m/s]
the velocity of object 2 at the instant AFTER the collision [m/s]
p
=
the momentum of the object [kg m/s]
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v2f
Case II: One or both objects are traveling in the positive OR negative direction
(i.e. backward) BEFORE the collision; one or both objects is traveling in the
positive OR negative direction AFTER the collision. In the case shown below,
both m1 and m2 are traveling to the right initially, only m2 is moving after the
collision. Note that a velocity in the negative direction is denoted by an “upsidedown” bucket.
After collision
Before collision
m1
v1i
v2i
m2
+
pm1
v1f
+
=
pm2
m1
-pm1
in
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pm2
m2
v2f
Case III: One or both objects are traveling in the positive OR negative direction
(i.e. backward) BEFORE the collision. However, after the collision, the two objects
lock/stick together and travel in the same direction (which may be positive or
negative). In the case shown below, both m1 and m2 are traveling to the right
initially, and after the collision they “lock” or “stick” together and move off with
a common velocity represented with vf.
After collision
Before collision
m1
v1i
v2i
m2
+
pm1
m1
=
pm2
pm1+m2
in
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m2
vf
Step 4: Write Conservation of Momentum Equation
Based on the buckets that were drawn for each case, the corresponding
conservation of momentum equation can be written. The “momentum buckets”
for each case are redrawn below for convenience.
Case I: Two objects are moving to the right before the collision. After the
collision, one object stops and the other continues to travel to the right.
+
m1v1i
+
=
m2v2i
m2v2f
=
Case II: Two objects are moving to the right before the collision. After the
collision, one object “rebounds” and travels to the left. The other object
continues to move to the right. Note the negative momentum for the objects
that “rebounds” and travels to the left.
+
m1v1i
+
m2v2i
=
+
=
-m1v1f + m2v2f
Case III: Two objects are moving to the right before the collision. After the
collision, the objects stick together and move off with a common velocity.
+
m1v1i
+
=
m2v2i
=
(m1+m2)vf
Step 5: Solve (the mathematical part)
After completing Step 4, the conservation of momentum equation for that
situation will be written. Given information for all but one variable, algebra can
be used to solve for the remaining variable. As mentioned above, this step is a
purely mathematical step, using math as a tool to arrive at the correct answer.
The following section models how this strategy can be used to solve a
Conservation of Momentum problem.
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Part III.
Example Conservation of Momentum Problem
Two trains are on a track. The mass of Train 1 is 1000 kg and the mass of Train 2 is
2500 kg. Train 1 is moving with a velocity of 20 m/s to the right. Train 2 is initially
at rest. The two trains collide and after the collision Train 1 continues to move to
the right but at 10 m/s. The trains do not couple with one another. (a) What is the
velocity of Train 2 after the collision? (b) If Train 1 instead rebounds with a
velocity of 5 m/s after the collision (i.e. moves backward), what is the velocity of
Train 2 after the collision, (c) If the trains DO couple with one another after the
collision, with what common velocity do they move off together with?
Given:
Train 1
v1i
Train 2
Train 1
v2i
v2f
Train 2
Find: a) vTrain2 after collision if Train 1 continues to move forward
b) vTrain2 after collision if Train 1 rebounds
c) vTrain1+2 after collision if Train 1 and Train 2 couple together upon collision
Solution:
PART A:
1. Statement of Conservation of Momentum
pi = pf
2. State positive direction
+
Positive direction is to the right
3. Draw Momentum Buckets
+
+
=
pTrain1
pTrain1
pTrain2
4. Write Conservation of Momentum equation
m1v1i
=
m1v1f + m2v2f
5. Solve
v2 f =
m1v1i ! m1v1 f
m2
=
(1000 )(20 ) ! (1000 )(10 ) = 4[m / s]
2500
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The positive answer implies the Train 2 travels in the positive direction, that is, to
the right.
PART B:
1. Statement of Conservation of Momentum
pi = pf
2. State positive direction
+
Positive direction is to the right
3. Draw Momentum Buckets
+
+
=
pTrain1
pTrain2
-pTrain1
4. Write Conservation of Momentum equation
m1v1i
=
-m1v1f
+ m2v2f
5. Solve
v2 f =
m1v1i + m1v1 f
m2
=
(1000 )(20 ) + (1000 )(10 )
2500
= 12[m / s ]
The positive answer implies the Train 2 travels in the positive direction, that is, to
the right. Note that train 2’s velocity is greater than in part (a). The fact that Train
1 “rebounds” implies that there was a greater force here than in part (a),
resulting in a greater change in momentum. This could be due to the physical
makeup (i.e. material) that the objects are made of.
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PART C:
1. Statement of Conservation of Momentum
pi = pf
2. State positive direction
Positive direction is to the right
+
3. Draw Momentum Buckets
+
=
pTrain1
pTrain1+ Train1
4. Write Conservation of Momentum equation
m1v1i
=
(m1+m2)vf
5. Solve
v2 f =
(1000 )(20 )
m1v1i
=
= 5.71[m / s ]
m1 + m2 (1000 + 2500)
The positive answer implies the Train 1+2 travel in the positive direction, that is, to
the right.
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