Download Modeling Collision force for carts Experiment 7

MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
Fall Term 2007
Problem Set 8: Simple Harmonic Oscillator and Collision Theory
Problem 1 (10 points): Dimensional Analysis, Estimation and Concepts
a) (2 points) Imagine that one drilled a hole with smooth sides straight through the center
of the earth, of radius Re = 6.4 ×106 m . If the air is removed from this tube (and the tube
doesn’t fill up with water, liquid rock or iron from the core), an object dropped into one
end will have enough energy to just exit the other end after an interval of time. Use
dimensional analysis to estimate that interval of time. Let g = 9.8 m ⋅ s −2 be the
gravitational constant.
b) (2 points) Estimate the energy loss in a completely inelastic collision between two
identical cars that collide head-on traveling at highway speeds.
c) (2 points) An explosion splits an object initially at rest into two pieces of unequal
mass. Which piece has the greater kinetic energy? Explain your answer.
1.
2.
3.
4.
The more massive piece.
The less massive piece.
They both have the same kinetic energy.
There is not enough information to tell.
d) (2 points) A ball is thrown against a wall; the ball bounces off and returns with speed
equal to the speed it had before striking the wall. Which of the following statements is
true from before to after the collision between the ball and the wall? Explain your answer.
1.
2.
3.
4.
5.
6.
The kinetic energy of the ball is the same.
The momentum of the ball is the same.
Both the kinetic energy and the momentum of the ball are the same.
Neither the kinetic energy nor the momentum of the ball are the same.
The collision is inelastic.
Two of the above.
e) (2 points) You just witnessed a karate master breaking a brick with his hand. Estimate
the impulse necessary to break the brick. Estimate an upper limit on the force per area
that the hand can safely endure without breaking the hand.
Solutions:
a) The only combination of the given parameters that has dimensions of time is
Re
= 810s = 13.5min .
g
(1.1)
If such a hole could be made, and the density of the earth were constant, the motion
would be simple harmonic. The force on an object would be its greatest at the surface of
the earth, decrease to zero as the object approached the center of the earth, and reverse
direction (that is, remain directed towards the center of the earth) after passing through
the center. For uniform density, the force would be proportional to the distance from the
center. If we model this motion as the object being subject to a restoring force with
“effective spring constant” keff , we have keff Re = mg , and the period of this motion
would be T = 2π Re / g = 87 min . It’s not hard to show (but not part of this problems)
that this is the period of a satellite in low orbit about the earth.
Equation (1.1) might be recognized as (within the factor of 2π , and with small
amplitude) as the period of a simple pendulum with length equal to the radius of the
earth. If anyone can construct such an item, many of us would pay to see it.
b) Consider small cars going fast, m = 1000 kg and v = 30 m ⋅ s −1 . The energy lost is the
sum of the initial kinetic energies,
2
1
2 mv 2 = (1000 kg ) ( 30 m ⋅ s −1 ) ≈ 106 J ≈ 1MJ .
2
(1.2)
c) The pieces have the same magnitude of momentum. The kinetic energy of an object of
mass m and momentum magnitude p is p 2 / 2m , so the piece with the smaller mass has
the larger kinetic energy. (This assumes that the pieces are not rotating, but that’s not a
consideration in this problem.)
d) 1): True – speed is given as unchanged. 2) Not true – momentum is a vector, and the
direction of the ball’s direction is reversed, and the momentum cannot be the same after
bouncing off the wall. The magnitude of the momentum is the same, but that’s not what
this question asks. 3), 4) Can’t be, since 1) is true and 2) is not. 5) Not true - if the speed
is the same, the collision is elastic. 6) Not true, since only 1) is true.
e) The motion is of course quite complicated. Taking the mass of the hand (should be the
hand and some part of the forearm, at least) as 1kg and the initial speed of the hand to be
10 m ⋅ s −1 , and assuming that all of the momentum of the hand is transferred to the brick
(probably not the case), the impulse that breaks the brick is about 10 kg ⋅ m ⋅ s −1 . If this
impulse, the change in the momentum of the hand, delivered over a time of roughly
1ms = 10−3 s , represents the largest force that the hand can withstand, that force is
roughly 104 kg ⋅ m ⋅ s −2 = 104 N . Using an area of 10cm 2 = 10−3 m 2 gives a force per area
of 107 N ⋅ m −2 . (The SI unit for force per area is the pascal, 1Pa = 1N ⋅ m −2 .
Problem 2 (10 points): Simple Harmonic Motion
Consider an ideal spring with spring constant k . The spring is attached to an object of
mass m that lies on a horizontal frictionless surface. The spring-mass system is
compressed a distance x0 from equilibrium and then released with an initial speed v0
toward the equilibrium position.
a) What is the period of oscillation for this system?
b) How long will it take for the object to first return to the equilibrium position?
c) What is the magnitude of the velocity of the object when it first returns to the
equilibrium position?
d) Draw a graph of the position and velocity of the mass as a function of time.
Carefully label your axes and clearly specify any special values.
When the spring reaches the equilibrium position, a wad of gum of mass mgum = (1/ 5) m
is dropped on the first object and sticks.
e) What is the new period of the system?
f) What is the maximum amplitude xmax the spring will stretch?
Solutions:
a) In the absence of friction, the system is a simple harmonic oscillator with period
T0 = 2π
m
.
k
(2.1)
b) Given that at t = 0 the position is x = x0 and the velocity is v = v0 , the position as a
function of time is
x(t ) = x0 cos ω 0t − ( v0 / ω 0 ) sin ω 0t
(2.2)
where ω0 = 2π /T0 and T0 is that found in part a), Equation (2.1). Note that v0 is given
as a speed, and hence must be positive (or zero). Setting x ( t1 ) = 0 in Equation (2.2)
yields
tan ω t1 =
⎛ x ⎞
x0
1
⇒ t1 = tan −1 ⎜ 0 ⎟
v0 / ω
ω
⎝ v0 / ω ⎠
(2.3)
where t1 is the first time after t = 0 at which the object passes the equilibrium position.
It’s important to note that in Equation (2.3), x0 and v0 are both positive, implying a
positive direction to the right in the figure. However, if the positive direction is taken to
be to the left, x0 would be negative, v0 would be positive, but Equation (2.2) would be
x(t ) = x0 cos ω0t + ( v0 / ω0 ) sin ω0t ,
(2.4)
leading to the same result. This is an indication that our choice of coordinate system
doesn’t matter.
It’s always best to check the limiting cases. If x0 = 0 , t1 = 0 ; the object starts at the
equilibrium position. If v0 = 0 , we can use the slightly improper form tan −1 ( ∞ ) = π / 2 to
obtain t1 = T0 / 4 , a quarter period.
c) There are many ways to find the speed v1 at the time t = t1 when the object passes
through the equilibrium position. Two will be presented here, one relying on calculus
and trigonometry and one relying on energy consideration. You make the call as to
which is “better.”
I – Calculus and Trig
Differentiating Equation (2.2) with respect to time and setting t = t1 as in Equation (2.3),
v ( t1 ) = − x0ω sin ωt1 − v0 cos ωt1
= − x0ω sin ⎡⎣ tan −1 ( x0ω / v0 ) ⎤⎦ − v0 cos ⎡⎣ tan −1 ( x0ω / v0 ) ⎤⎦ .
(2.5)
At this point, we use the trigonometric relations
cos ⎡⎣ tan −1 α ⎤⎦ =
1
1+ α 2
, sin ⎡⎣ tan −1 α ⎤⎦ =
α
1+ α 2
(2.6)
to obtain for v1 = v ( t1 ) , after a bit of algebra,
v1 =
x02ω 2 / v0
1+ x ω / v
2
0
2
2
0
+
v0
1 + x02ω 2 / v02
(2.7)
= x02ω 2 + v02 .
(To see a graphical representation of the formulas in (2.6), draw a right triangle with one
side of α , the other side 1 and a hypotenuse 1 + α 2 .)
II – Energy Considerations
The initial mechanical energy is the sum of the kinetic and potential energies,
1
1
E0 = mv02 + kx02
2
2
(2.8)
and the mechanical energy E1 when the object passes through the equilibrium point is
1
E1 = mv12 .
2
Setting E0 = E1 and ω 2 = k / m reproduces the result in (2.7).
d) In the plot shown, x0 = 1 , v0 = 1 , k = 1 , m = 1 , so ω = 1 scaled appropriately.
(2.9)
The period is T0 = 2πω = 2π . From Equation (2.3), t1 = π / 4 . From (2.7), v1 = 2 > v0 .
Although not part of this problem, the amplitude is
2 > x0 . Maybe you can tell from
this plot that v1 > v0 .
e) The mass is now m2 = ( 6 / 5 ) m , and the period is T2 = 2π
( 6 / 5 ) m / g > T0 .
f) Of the many ways to find the new amplitude, energy and momentum considerations
might be the easiest to use. After the collision (the gum has stuck to the block), the speed
of the block-gum combination will be
m v1
5
= v1 ,
m + mgum 6
(2.10)
and so the mechanical energy has decreased by a factor of 5 / 6 . The new amplitude will
decrease by a factor of
5 / 6 , xmax = 5 / 6 x02 + ( v1 / ω ) .
2
Problem 3 (10 points): Two- Dimensional Particle Collision
Particle 1 of mass m1 is initially moving in the positive x -direction (to the right in the
figure) with a speed v1,0 and collides with a second particle 2 of mass m2 = m1 / 3 , which
is initially moving in the opposite direction (to the left in the figure) with an unknown
speed v2,0 . Assume that the total external force acting on the particles is zero. You do
not know whether or not the collision is elastic. After the collision particle 1 moves
with a speed v1, f = v1,0 / 2 , at an angle θ1, f = 90o with respect to the positive x -direction
(downward in the figure). After the collision, particle 2 moves with an unknown
speed v2, f , at an angle θ 2, f = 45o with respect to the positive x -direction (upward and to
the right in the figure). In this problem you will find v2,0 and v2, f in terms of v1,0 . You
also will determine whether or not the collision is elastic. Note: sin 45o = cos 45o = 2 / 2 .
a) Find the speed v2,0 in terms of v1,0 .
b) Find the speed v2, f in terms of v1,0 .
c) Is the mechanical energy conserved in this collision? Justify your answer.
Solution:
a) The problem statement specifies that there are no net external forces acting on the
system. Therefore, momentum is conserved.
In the above diagram, take the positive î -direction to be to the right and the positive ˆj direction to be upward.
To conserve momentum in the î -direction,
m1v1, 0 − m2 v2, 0 = m2 v2, f
v1, 0 −
v2, 0
3
=
v2, f
2
2
(3.1)
,
3 2
where m2 = m1 / 3 and the known trigonometric relations have been used.
To conserve momentum in the ĵ -direction,
2
− m1v1, f
2
v2, f v1, 0
0=
−
,
2
3 2
0 = m2 v2, f
(3.2)
where the same relations have been used.
A simple substitution of the second expression in (3.2) into the second expression in (3.1)
gives v1, 0 = −v2, 0 / 3 = v1, 0 / 2 , readily solved for
v2, 0 =
3
v .
2 1, 0
(3.3)
(
)
b) The second expression in (3.2) immediately gives v2, f = 3/ 2 v1,0 .
c) The initial and final kinetic energies are
E0 =
1
1
1
1 m1 ⎛ 3 2 ⎞ 7
m1v1,2 0 + m2 v2,2 0 = m1v1,2 0 +
v1, 0 ⎟ = m1v1,2 0
⎜
2
2
2
2 3 ⎝2 ⎠ 8
2
⎞
1
1
1 ⎛ v1, 0 ⎞
1 m1 ⎛ 3
7
E f = m1v1,2 f + m2 v2,2 f = m1 ⎜
v1, 0 ⎟ = m1v1,2 0 .
⎟ +
⎜
2
2
2 ⎝ 2 ⎠
2 3⎝ 2
8
⎠
2
(3.4)
The energies before and after the collision are the same; mechanical energy is conserved.
Problem 4 (10 points): Experiment 4 Modeling the Elastic Collision
Part a) (5 points) The magnitude of the repulsive force in Experiment 4: Momentum and
Collisions between the two carts is modeled by the force law
F = F0 e − x / l
(4.1)
where x is the separation between the carts and l is a characteristic length associated
with the magnets in the carts. If the carts start very far apart (an infinite distance), the
work required to bring them to a separation distance d is:
d
d
∞
∞
W = − ∫ F ⋅ dr = − F0 ∫ e− x / l dx .
(4.2)
Calculate this integral. The result is the equivalent of the potential energy stored in a
⎛1
⎞
spring ⎜ k x 2 ⎟ that satisfies a linear restoring force law.
⎝2
⎠
Part b) (5 points) Suppose there are two of these carts on a track. One cart with mass m
and initial speed v A,1 runs into the other, which has equal mass and is initially at rest on
the track. Find the minimum separation distance d between the carts during the
collision. You may assume that F0 is sufficiently large so that d > 0 (the carts do not
make physical contact). Evaluate your result numerically using F0 = 2.0 × 101 N ,
l = 5.0 × 10 −3 m , m = 2.5 × 10 −1 kg and v A,1 = 5.0 ×10−1 m ⋅ s −1 . Hint: This problem is
most easily done in the reference frame that is moving at a velocity equal to the center of
mass velocity of the two-cart system. This frame is called the center of mass
reference frame.
Solutions:
a) The improper definite integral is straightforward;
d
d
∞
∞
W = − F0 ∫ e − x / l dx = F0 l ⎡⎣ e − x / l ⎤⎦ = F0 l e − d / l
(4.3)
(be sure to recognize that the upper limit, the distance “ d ,” is not the same as the
differential in “ dx .”) The exponential factor e − d / l is dimensionless, so the result in (4.3)
has dimensions of force times distance, the dimensions of energy.
b) In the center of mass frame, each cart has speed v A,1 / 2 , and the initial kinetic energy
2
1 ⎛v ⎞
in this frame is 2 m ⎜ A,1 ⎟ = m v A2 ,1 / 4 .
2 ⎝ 2 ⎠
In this frame, when the separation is a
minimum, both carts will be stationary, with no kinetic energy, and the total mechanical
energy in this frame is F0 l e − d / l . Setting
m v A2 ,1
4
= F0 l e − d / l
(4.4)
and solving for d yields, after some basic math,
⎡ 4F l ⎤
d = l ln ⎢ 02 ⎥ .
⎢⎣ m v A,1 ⎥⎦
(4.5)
This result makes sense: If F0 is larger, the overall repulsive force is larger, and the
minimum separation is larger. If v A,1 is larger, the minimum separation is smaller. What
might not be intuitive is that if the characteristic length l is larger, the minimum distance
is larger. A larger l would mean a stronger force at larger distances, hence more
repulsion between the carts and a larger minimum separation.
Problem 5 (10 points): Experiment 4 Momentum and Collisions Analysis
Part One: Inelastic Collision: In Experiment 4, you recorded data for the initial and final
velocities of the incident and target carts for inelastic collisions. Complete the analysis of
your data from the inelastic collisions by following the two steps below, and answer parts
a) and b) for each step.
You will analyze this collision as seen by two different observers. One observer is at rest
with respect to the classroom; this reference frame is called the lab reference frame. The
other observer is moving in the center of mass reference frame.
Step 1 (1 point): Draw two momentum diagrams for the carts immediately before the
collision and immediately after the collision, one describing the collision in the lab
reference frame, and one describing the collision in the center of mass reference frame.
Analysis:
Step 2: You made three different trial collisions; the first with equal masses, and the
second and third with unequal masses. Use your measured values to complete Table 1.
Entries to the left of the double vertical line are before the two carts collide; those to the
right are after the two carts collide but before the collision with the force sensor. Hand
the table in with your homework solutions. You do not need to include standard
deviations or error estimates but you may want to consider them when answering part b)
below.
Part a) (2 points): First calculate the initial kinetic energy K A,1 and the final kinetic
energy K 2 in the lab reference frame. The difference between the two, K 2 − K A,1 = Wnc ,
is the non-conservative internal work done during the totally inelastic collision.
Part b) (2 points): Then calculate the velocity of the center of mass, which is given by
v cm =
mA v A,1
.
mA + mB
(5.1)
Because momentum is conserved, you should expect v cm = v 2 . Calculate the total kinetic
energy of the two carts in the center of mass reference frame given by
K cm,1 =
2
1
1
2
mA ( v A,1 − vcm ) + mB ( vcm ) .
2
2
(5.2)
In the center of mass reference frame the incident cart A moves forward with a slower
velocity, v A,1 − v cm , than in the lab frame; the cart B moves backward with a velocity
− v cm . After the collision the two carts are stuck together and at rest in the center of mass
frame, so the final kinetic energy in this frame is K cm,2 = 0 . Therefore the nonconservative work in the center of mass reference frame is given by
Wnc = K cm, 2 − K cm, 1 = − K cm, 1 .
(5.3)
Show that K 2 − K A,1 = − K cm, 1 . This will require a calculation. Do your experimental data
verify this theoretical result?
Part Two: Elastic Collision: Complete the analysis of your data for the elastic collisions
by following the two steps below, and parts a) and b) below.
Once again you made three trials. Since the motion sensor can only measure the initial
and final velocities of the incident cart A, you needed to calculate the impulse of the
target cart B on the force sensor to determine the final velocity of cart B.
Step 1 (1 points): Draw two momentum diagrams for the carts immediately before the
collision and immediately after the collision, one describing the collision in the lab
reference frame, and one describing the collision in the center of mass reference frame.
Analysis:
Step 2: If you assume that the cart B collides elastically with the force sensor, then
during the collision the momentum of the cart B changes by Δp B = −2mB v B , 2 . This
change in momentum is the impulse that the force sensor exerts on the target cart B. Cart
B therefore exerts an equal and opposite impulse on the force sensor; both have
magnitude I = 2mB v B,2 . Measuring the impulse allows you to calculate the velocity of the
target cart B after the collision according to
v B,2 = I / 2mB
(5.4)
You made three different trial collisions: the first with equal masses, and the second and
third with unequal masses. Use your measured values to complete Table 2. Entries to the
left of the double vertical line (upper part of the table as shown below) are before the two
carts collide; those to the right (lower part of the table) are after the two carts collide but
before the collision with the force sensor. You do not need to include standard deviations
or error estimates but you may want to consider them when answering the question
below.
Question (4 points): Assume that the collision between the carts is elastic. Use the
principles of conservation of momentum and mechanical energy to calculate the
theoretical post-collision velocities, v A, 2 for cart A and vB , 2 for cart B, for each of the
three trials. Show your work. (These calculations are more complicated for the cases
where the carts have different masses.)
Enter the results in Table 3. The values to the left of the double vertical line (upper part
of the table as shown below) are your measured values from Table 2, those to the right
(lower part of the table) are calculated theoretical values.
Briefly describe whether or not your measured values agree with your theoretical values.
Do your results confirm that the collisions are elastic?
Solutions:
These solutions use the data from the Example in
http://web.mit.edu/8.01t/www/materials/Experiments/examples/exp04ex.htm
The completed tables below were generated using an Xcel spreadsheet. Data are
represented by red, calculations in black. Note that not all calculations were rounded to
the proper precision.
Part One: Inelastic Collision:
Step 1: The arrows representing the velocities are not necessarily to scale.
Step 2:
Table 1 – Inelastic Collisions
mA
0.25 kg
0.25 kg
0.50 kg
mB
0.25
0.50
0.25
vA,1 ( m ⋅ s −1 ) vcm ( m ⋅ s −1 )
B
kg
kg
kg
0.6935
0.9523
0.5204
0.347
0.317
0.347
KA,1 (J)
0.060
0.113
0.068
Analysis:
The initial kinetic energy K cm, 1 in the center of mass frame is
Kcm,1 (J)
0.030
0.076
0.023
v2 ( m ⋅ s −1 )
0.3369
0.3105
0.3374
K2 (J)
0.028
0.036
0.043
Wnc (J)
-0.032
-0.077
-0.025
2
1
1
2
K cm, 1 = mA ( v A,1 − vcm ) + mB vcm
2
2
2
⎛ mB ⎞ 1
1
mA ⎞
2 ⎛
= mAv A2 ,1 ⎜
⎟
⎟ + mB v A,1 ⎜
2
⎝ mA + mB ⎠ 2
⎝ mA + mB ⎠
=
2
(5.5)
1 mAmB 2
v A,1
2 mA + mB
where vcm = v A,1mA / ( mA + mB ) has been used. The final kinetic energy in the lab frame is
1
1
( mA + mB ) vcm2 = vA2 ,1mA2 / ( mA + mB ) , and so the change in kinetic energy in the lab
2
2
frame is
K2 =
1 2
mA2
1
− v A2 ,1mA
K 2 − K A,1 = v A,1
2
mA + mB 2
⎛ mA
⎞
1
= v A2 ,1mA ⎜
− 1⎟
2
⎝ mA + mB ⎠
= − K cm,1.
(5.6)
This is consistent with the experimental results.
Elastic Collisions:
The figures are on the next page. As with the inelastic collisions, the length of the arrows
is schematic only (but the directions are what we expect).
Question 4:
A completely elastic collision in one dimension is most easily analyzed by considering
the collision in the center of mass frame. As in the inelastic collision analysis, the
velocity this frame (the component of velocity in the direction of the original motion of
the first cart) is vcm = v A,1mA / ( mA + mB ) , cart A has an initial velocity v A,1 − vcm and cart
B has an initial velocity of −vcm . The only way to conserve kinetic energy and maintain
zero momentum in the center of mass frame is if the velocity of each cart changes
direction but keeps the same magnitude, or vcm − v A,1 and −vcm respectively. The final
velocities of the carts are then
v A, 2 = vcm + ( vcm − v A,1 ) = v A,1
m − mB
,
= v A,1 A
mA + mB
2mB
mA + mB
(5.7)
vB ,2 = vcm + vcm = v A,1
2m A
.
mA + mB
Table 2 – Elastic Collisions
vA,1 ( m ⋅ s −1 )
mB
mA
0.4865
0.25 kg 0.25 kg
0.5967
0.25 kg 0.75 kg
0.3747
0.75 kg 0.25 kg
B
vcm ( m ⋅ s −1 )
0.24325
0.149175
0.281025
KA,1 (J)
0.059170563
0.044506361
0.052650034
(Table 2 Continued)
vA,2 ( m ⋅ s −1 )
-0.009004
-0.2796
0.174
I kg ⋅ m ⋅ s −1
0.2353
0.4309
0.2771
vB,2 ( m ⋅ s −1 )
0.471
0.287
0.554
KB,2 (J)
0.027683045
0.030945802
0.038392205
(5.8)
Table 3: Elastic Collision Velocities, Measured vs. Theoretical Values
Measured Values
mA
0.25
0.25
0.75
mB
0.25
0.75
0.25
B
kg
kg
kg
kg
kg
kg
Theoretical Values
vA,2 ( m ⋅ s −1 )
vB,2 ( m ⋅ s −1 )
0
-0.29835
0.18735
0.4865
0.29835
0.56205
The agreement is quite good.
vA,2 ( m ⋅ s −1 )
vB,2 ( m ⋅ s −1 )
-0.009004
-0.2796
0.174
0.471
0.287
0.554