Download AP50 Fall 2016 Problem Set 2 Solutions 1) Reindeer crossing

AP50 Fall 2016
Problem Set 2 Solutions
1) Reindeer crossing: Jackie is driving on a highway road at night under very icy conditions. She
drives over a stretch of iced-over road and loses control of the car. Beyond her headlights, she sees a
stationary reindeer with which she eventually collides. What is the final velocity of the reindeer?
What is the final velocity of the car? What is the difference in the total momentum of the deer+car
before and after the collision? (Make reasonable estimates of relevant quantities) Answer units: [m/
s]
I: Getting Started
Jackie is driving (before the collision) on a highway road and is moving at a typical highway speed.
She drives onto an icy patch and loses control of her car. Because she loses control of the car we can
assume she is slipping on the ice and can approximate her motion as one on a frictionless surface.
She is thus sliding on the ice at the same velocity at which she was previously driving. She slides
toward and hits the stationary reindeer. Both her and the reindeer are on ice, so we can take the car
+deer system to be isolated since friction, the only external force acting on the car+deer system, is
not present. But because the car and reindeer collision is likely inelastic (because both car and
reindeer are deformable) we have to introduce a coefficient of restitution into our calculation.
The objective is to determine three pieces of information: The velocity of Jackie after the collision;
the velocity of the reindeer after the collision; whether the system conserves total momentum.
II: Devise Plan
The plan needed to achieve this objective is
1) Estimate values for the inertia of Jackie, car, and deer; estimate the velocity of Jackie before
collision; estimate coefficient of restitution.
2) Assume system collision proceeds in one dimension.
3) Write momentum conservation and relative velocity equations (with coefficient of restitution)
4) Solve equations for final car velocity and final deer velocity
5) Use physical principles or direct calculation to determine whether momentum is conserved.
III: Execute Plan
- The inertia of a car is about 1000 kg (http://hypertextbook.com/facts/2000/YanaZorina.shtml).
- The inertia of the average woman is about 60 kg (https://en.wikipedia.org/wiki/
Human_body_weight).
- The inertia of a reindeer is about 180 kg (https://en.wikipedia.org/wiki/Reindeer)
- The collision is inelastic (because the car is deformable and the reindeer is “squishy”) so we
should introduce a coefficient of restitution. Given that the coefficient of restitution between a
glass marble and wood (a system which should be more energy conserving than our own) is e
=0.5 (http://www.quintic.com/education/case_studies/coefficient_restitution.htm), we can take e =
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0.5 to be an upper limit for our collision. All of our subsequent velocity results will be upper
limits.
- On a highway we can expect a car to be traveling at 65 mph or (equivalently) 30 m/s. Denoting
the car + Jackie system as C for simplicity and the reindeer as D we thus have
mC = 1000 kg + 60 kg = 1060 kg
mD = 180 kg
vCi = 30 m/s
e = 0.5
- If we take the system to be moving in one dimension, we can depict the system before and after
collisions as
- To find v_{Cf} the car’s (and Jackie’s) final velocity and v_{Df} the reeindeer’s final velocity, we
apply the momentum conservation formula
mC vCi = mC vCf + mD vDf
and the “coefficient-of-restitution” corrected relative velocity formula
vCf
vDf =
e (vCi
vCf
vDf =
evCi
vDi )
Solving for v_{Cf} in the relative velocity formula and plugging the result into the momentum of
conservation equation we find
mC vCi = mC (vDf
evCi ) + mD vDf
(mC + emC )vCi = (mC + mD )vDf
!
vDf =
(1 + e)mC
vCi
mC + mD
And using this result in the coefficient of restitution formula gives us for v_{Cf}
vCf = vDf
evCi
mc (1 + e) e(mc + mD )
vCi
mC + mD
mC emD
=
vCi
mC + mD
=
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Using our previously defined parameter values we thus obtain
vDf =
vCf
✓
1060 kg(1 + 0.5)
1060 kg + 180 kg
◆
⇥ 30 m/s
= 38.47 m/s
✓
◆
1060 kg (0.5)180 kg
=
⇥ 30 m/s
1060 kg + 180 kg
= 23.45 m/s
These values are upper limits to the velocity because we are using a coefficient of restitution of e =
0.5 (which is itself an upper limit). Because the system is isolated (i.e., there are no external forces
acting upon it), the total momentum is conserved and the momentum before the collision is
equal to the momentum after the collision.
IV: Evaluate Result
There are a number of ways to evaluate or extend this result. We can check our conclusion that the
total momentum is conserved by computing the momentum of our system both before and after the
collision.
Before the collision our total momentum is
Ptot,i = mC vCi = (1060 kg)(30 m/s)
= 3.18 ⇥ 104 kg m/s
After the collision our total momentum is
Ptot,f = mC vCf + mD vDf = (1060 kg)(23.45 m/s) + (180 kg)(38.47 m/s)
= 3.18 ⇥ 104 kg m/s
which matches the momentum before the collision up to our rounded place value. We should note
that for isolated systems, momentum conservation always applies and the momentum before the
collision would equal the momentum after the collision regardless of the value of the coefficient of
restitution.
We note as well that if we consider our collision to be completely inelastic and set e = 0 in our
algebraic formulas for the velocity, then the car and the reindeer move with the same final velocity
suggesting they are stuck together, as we expect for a completely inelastic collision.
AP50 Fall 2016
2) Bumper Carts. Two carts are initially moving to the right on a low-friction track, with cart 1
behind cart 2. Cart 1 has a speed twice that of cart 2 and so moves up and rear-ends cart 2, which
has twice the inertia of cart 1. What is the speed of each cart right after the collision if the collision
is elastic?
I: Getting started:
We can begin by organizing the information into a sketch to help visualize the situation. We need to
determine the velocities after the collision from the information given, knowing that the collision is
elastic.
!
II: Devise a Plan:
To determine the two unknowns v1x,f and v2x,f, we need two equations. Because the collision is
elastic we can use Eqs. 5.4 and 5.5 from the course textbook.
III: Execute Plan:
We begin by solving equation (1) for v2x,f :
so that we can eliminate v2x,f in equation (2):
Solving this expression for v1x,f yields
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If we now plug in m2= 2m1 and v2x,i= (0.5) v1x,f which we know from the problem statement we
have:
Repeating this procedure, we can eliminate v1x,f and solve for v2x,f:
We can plug in m2= 2m1 and v2x,i= (0.5) v1x,f.
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IV: Evaluate Result:
The final velocities yield the final relative speed:
This value is the same as the initial relative speed:
as required for an elastic collision.
Also, the momentum before the collision is:
while the momentum after the collision is:
equal the momentum before.
These two checks show that no algebraic errors were made, since these are the requirements of the
two equations we used.
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3) The Big Bang. Suppose an ice block of inertia 6.0 kg is moving in one dimension with a speed of
3.0 m/s across a slick, horizontal sheet of steel which we assume to have negligible friction. The ice
block suddenly breaks into one fragment of inertia 2.0 kg and another of inertia 4.0 kg. Suppose the
velocity of the 2.0 kg piece is 2.0 m/s, along the initial direction of motion. (a) If the velocity of the
4.0 kg piece is 3.5 m/s, also along the initial direction of motion, what is the momentum the system
comprising the two fragments before and after the fracture? Is the momentum of the system
constant? (b) What is the kinetic energy of the system before and after the ice breaks? Is the
mechanical energy of the system constant? Why or why not? (c) What is the center-of-mass velocity
before and after the ice breaks? Does the center-of-mass velocity change?
I: Getting Started:
First, we organize all of the information given in the problem by drawing it pictorially.
!
!
II: Devise Plan:
We can find the momentum before and after the ice breaks, using px = mvx and the given inertias and
velocities. The system is isolated so we know momentum will be constant. The kinetic energy of the
system can be found using K = ½ mv2. The center of mass velocity can be calculated using eqn.
6.26.
!
III: Execute Plan:
a) We use the given velocities and inertias to calculate the momentum before and after the ice
breaks.
!
We see that yes, as we expect, the momentum is the same both before and after the fracture
in this isolated system.
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b) We calculate the kinetic energy before and after the ice breaks and see that it actually
increases. Therefore, the mechanical energy of the system is not constant, meaning there is
some change in internal energy of the system.
!
c) Finally, we calculate the center-of-mass velocity. Since all pieces of ice are always traveling
along the initial direction of motion along the +x axis, the direction of the velocity vector is
to the right along x, or in the direction.
!
IV: Evaluate Result:
We have an isolated system, so momentum is conserved as we expect in part a.
In part b, we calculated that the kinetic energy increases, but where is it coming from? There must
be some internal energy that is transferred to mechanical energy when the ice block breaks apart. If
we look back to sections 5.6 and 5.8 of the text on explosive separations we see that when this
happens, the final relative velocity is greater than the initial relative velocity. In this problem, the
two pieces are stuck together initially moving at the same velocity, so v12i = 0 but later they move at
different velocities so v12f > 0 meaning that our coefficient of restitution e > 1 and this does look
like an explosive separation as we expect from our sketch.
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In part c, all pieces are moving to the right along our x-axis so we expect our center of mass
velocities both before and after separation to also be in that direction, as they are. AP50 Fall 2016
4) Waterfall. What is the magnitude of the momentum change of 2 gallons of water (inertia about
7.6 kg) as it comes to a stop in a bathtub into which it is poured from a height of 2.0m?
I: Getting Started:
Before we start, we sketch out the situation and then convert it into a motion diagram.
!
We want to find how much the momentum of the water changes as it comes to rest after being
poured into a bathtub. We will assume that the water is poured slowly so that it has negligible
velocity as it starts to fall from the container that it is poured from. This also means we are
assuming a one dimensional vertical, free-fall situation with no initial velocity and a constant
acceleration of –g on the water due to gravity. We want to know how much the momentum changes
as the water comes to a stop in the bathtub. The water has its highest velocity and therefore highest
momentum right before it hits the bathtub where it begins to slow down and eventually come to
rest. We need to find the velocity it has when it reaches the tub to calculate the momentum change.
II: Devise Plan:
We can use the constant acceleration or free-fall motion equations from section 3.6 to determine the
final velocity of the water just before it reaches the tub.
!
After we have the final velocity of the water, we will multiply it by the inertia of the water to find
out the momentum it has as it reaches the tub.
III: Execute Plan:
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First, we use eq. 3.13 to find the velocity of the water just before it reaches the tub.
!
We take the negative value for as its direction is downward. Plugging in the values given in the
problem,
Now, we want to find the magnitude of the change in momentum of the water as it comes to a stop
in the bathtub. We know that inertia is always positive, but since velocity is a vector, it can be
positive or negative. We are interested in the magnitude of the change, so we take the magnitude of
the difference in velocity.
!
In this case, we are interested in the momentum change “as it comes to a stop in a bathtub”.
Therefore, we define our initial velocity as that of the water just before it hits the tub and the final
velocity as 0m/s (when it comes to a stop in the tub) giving us a final magnitude of momentum
change of 48 kg·m/s.
!
IV: Evaluate Result:
Two gallons of water at 7.6kg is a reasonable inertia since a liter of water has inertia 1kg, and a
gallon has just under 4 liters, so we would expect 2 gallons to have just under 8kg.
6.3 m/s seems to be a reasonable speed for water poured from 2 meters to be traveling, so we get
also get a reasonable number for the momentum change. If we look at the equations we used, we
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see that if we increase the inertia or the speed of the water, the change in momentum increases as
we expect. We see that we can increase the speed by increasing the height we pour the water from
or by increasing the acceleration with which it falls, also as we expect.
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5) Fly Ball. Revisit the bouncing balls EDA from class: ball 1 with inertia m1 is placed on top of
ball 2 with inertia m2 >> m1. The combination is dropped from a height h above a hard surface. In
class, you assumed all collisions were elastic and predicted the final height of the small ball, but
then found the ball never reached this height. (a) How would the predicted height change if you
allowed the collisions to be inelastic, with coefficient of restitution e? Assume both collisions have
the same coefficient of restitution and that the radii of the balls are much smaller than the height h
from which they are dropped. (b) Modify your solution taking into account two different
coefficients of restitution for the two successive collisions. (c) Estimate the coefficient(s) of
restitution from your data of the bouncing balls video from the EDA.
I: Getting Started:
We first draw the problem to organize the information:
!
II: Devise Plan:
(1) Compute the velocity, v, of both balls just before impact using
v2 = vi2 + 2a(Δh), where v is the velocity of the balls before hitting the ground (same as the
velocity of ball 1 at that time) and vi is 0, the velocity of the balls when they are dropped.
(2) Compute the velocity of ball 2 after it interacts with the ground, assuming coefficient of
restitution e2g.
(3) Compute the velocity of ball 1 after it interacts with ball 2, assuming m2>>m1 and a
coefficient of restitution e12.
(4) With the velocity of ball 1 after the collisions, compute the final height hf, using the equation
in step 1. This will be expressed in terms of the initial height h and the coefficient of
restitution.
(5) Using this equation, plug in the final height from the video and solve for the coefficients of
restitution.
AP50 Fall 2016
III: Execute Plan:
(1) The balls are dropped with zero initial velocity. We define positive to be upwards, so the
acceleration is –g. The change of position is the final position minus the initial position, or
0−h = −h.
1.
Rearranging the equation in the plan (step 1), we have that the speed of the balls just
before they hit the ground is v = sqrt(2gh), since the negatives cancel, and the direction is
downwards.
(2) Ball 2 undergoes an inelastic collision with the ground, with coefficient of restitution e2g.
!
The initial relative speed between the ball and the ground is v, so the final relative speed is
e2g*v. Thus, ball 2 leaves the ground at an upward velocity of e2g*v.
(3) Now, ball 2, heading upward at e2g*v, interacts with ball 1, heading downward at v, as
shown in figure below:
!
Their relative speed before the collision is therefore e2g*v + v = v(1+ e2g). Since this is an
inelastic collision with coefficient of restitution e12, the relative speed after the collision is
e12*v(1+ e2g).
Since m2>>m1, we can assume that ball 2’s velocity does not change during and after the
collision. Therefore, the velocity of ball 1 after the collision is the velocity of ball 2 plus the
relative speed between the two balls: e2g*v + (1+ e2g) e12*v= v*(e2g+e2ge12+e12).
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(4) Using the equation from part 1, we solve for the height at which the final velocity is zero.
Thus,
0=[v(e2g+e2ge12+e12)]2 + 2(−g)hf
hf = v2(e2g+e2ge12+e12)2/ 2g .
since v = sqrt(2gh), from step 1.
For part (a), we assume e2g = e12 = e and use v = sqrt(2gh), from step 1. The maximum
height of the ball is therefore
hf = (e2+2e)2h
or (e2+2e)2 times the height it was dropped from.
For part (b), e2g and e12 are not the same but (v = sqrt(2gh) is still true) the maximum height
of the ball is therefore
hf = (e2g+e2ge12+e12)2 h
(e2g+e2ge12+e12)2 times the height it was dropped from.
(5) In the video, the ball went about 2.8 times higher than it was dropped. We make the
approximation that the coefficients of restitution for the two successive collisions are the
same. This works out to
(6)
e2+2e−sqrt(2.8)=0
which when solved for non-extraneous solutions yields e = 0.64
We could alternatively use two videos (two observations) to compute e2g and e12 separately.
We first use the video of ball 2 falling on its own to compute e2g, then the video of both balls
falling together to compute e12.
For ball 2 falling on its own, we use steps 1-2 above to determine that its initial velocity
after bouncing is e2g*v. We then want to find how high it goes. We use the kinematics
equation from step 1 to solve for the time when the velocity is zero, since this is the
maximum height. Thus, 0 = (e2g v)2 + 2(−g)hf1, but v = sqrt(2gh) (from step 1), so we get
hf1 = e2g2 * h.
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In the video, we observe that ball 2 bounces to 70% its original height, so
hf1/h = e2g2 = 0.70, so e2g = 0.84.
To solve for e12 from the video with two balls, we find that ball 1 goes 2.8 times higher than
it was dropped, then we use the equation from the previous step:
hf /h = 2.8 = (e2g+e2ge12+e12)2.
Solving for e12, we get e12 = 0.45.
IV: Evaluate Result:
We first check the limits: what happens when e=1 and e=0? When e=1, corresponding to elastic
collisions, the ball goes 9 times higher than when it was dropped, which is what we computed in the
EDA. If e=0, we have a completely inelastic collision: the first ball sticks to the floor, and the
second ball sticks to the first, so the final height should be zero. Indeed, it is. Coefficients of
restitution between 0 and 1 give intermediate heights, as expected. [TF note: Checking limit is a
powerful tool to confirm the correctness when solving physics problems.]
All the coefficients of restitution came out between 0 and 1, as expected. However, e12=0.45 is a
relatively low coefficient of restitution – this would mean that ball 1 dropped on a stationary ball 2
would only bounce to 20% of its original height (using hf1 = e2g2 * h from the last step of the
solutions).
The coefficient of restitution relates to kinetic energy losses during an inelastic collision. Since we
are not considering any other sources of kinetic energy loss, any kinetic energy losses in the video
are all being attributed to the collisions. If kinetic energy is dissipated in other ways, for example
through air resistance, the inelasticity of the collision would be responsible for less energy
dissipation and our estimated coefficient of restitution would rise as a result. We could estimate the
coefficient of restitution in a different way in order to eliminate the effects of air resistance by using
the velocities just before and just after the collisions.
We also notice that ball 2 only bounces to 45% of its original height when ball 1 is on top, but to
70% of its original height when bouncing on its own. We assumed the mass of ball 2 was much
greater than that of ball 1, so ball 1 would not affect the velocity of ball 2 during their collision – i.e.
ball 2 should still bounce to 70% of its original height with ball 1 on top. This indicates that the
masses are too similar to ignore the effects of ball 1 on ball 2, and also explains why ball 1 did not
go higher and why e12 ended up being so low.