8.01L: Physics I November 7, 2015 Prof. Alan Guth
Problem Set #8 Solutions
Due by 11:00 am on Friday, November 6 in the bins at the intersection of Buildings 8 (3rd floor) and 16 (4th floor). Please put your name and recitation number clearly on the top page of your problem set.
SOURCES:
University Physics, Volume 1, 13th Edition, by Hugh D. Young and Roger A. Freed- man (Addison-Wesley, 2012). This is the required textbook for the course.
Essentials of Introductory Classical Mechanics, 6th Edition, by Wit Busza, Susan Cartwright, and Alan H. Guth, available to the MIT community athttps://web.mit.edu/
8.01L/studyguide/index.shtml. Also known as the BCG Study Guide.
READING: Young and Freedman Chapter 7, Sections 4–5 (POTENTIAL ENERGY AND ENERGY CONSERVATION: Force and Potential Energy; Energy Diagrams) and Chapter 8, Sections 1–2 (MOMENTUM, IMPULSE, AND COLLISIONS: Momentum and Impulse;
Conservation of Momentum).
OPTIONAL ADDITIONAL READING: Busza, Cartwright, and Guth (BCG), Chap- ter 5, pp. 164–172. (“Optional” means that you should read this only if you find it useful.) NOTE: Your written solutions must include a brief commentary in addition to any equations or graphs used to arrive at your answer. For example, this commentary should explain your basic strategy for solving the problem and also highlight the concept before applying an equation. We also want you to first solve a problem algebraically before you put in the particular numbers relevant to the problem. We’ll say that again: solve the answer in terms of variables before you start sticking in numbers! This makes it easier for you when you check your work and also for the grader to follow your logic.
8-1: Brake Failure
Young and Freedman (13th edition) problem 7.66, page 238.
(same as Young and Freedman (12th edition) problem 7.66, page 244.)
Fig. P7.66:
A truck with mass m has a brake failure while going down an icy mountain road of constant downward slope angle α (Fig. P7.66). Initially the truck is moving downhill at
speed v0. After careening downhill a distance L with negligible friction, the truck driver steers the runaway vehicle onto a runaway truck ramp of constant upward slope angle β.
The truck ramp has a soft sand surface for which the coefficient of rolling friction is µr. What is the distance that the truck moves up the ramp before coming to a halt? Solve using energy methods.
Solution:
Let the distance on the truck ramp be d. Also, let’s define the point A as the starting point at which the truck is a distance L from the truck ramp and its velocity is v0, and B the point at which the truck stops. We use energy transformation law on the truck+earth system:
∆E = Wnon−cons (1)
EB− EA = f~k · ∆~r (2)
EB− EA = −µkN d where N = mg cos β (3) Now:
EA = KA+ UA= 1
2mv02+ mgL sin α (4)
EB = KB+ UB = mgd sin β (5)
since the velocity at point B is zero. Using equation 3
EB− EA = −µkmgd cos β (6)
So, from equations 4 and 5:
mgd sin β − mgL sin α − 1
2mv02 = −µkmgd cos β (7) gd(sin β + µ cos β) = gL sin α +1
2v20 (8)
d = L sin α + 12v02/g
sin β + µ cos β (9)
Note: as expected, d = L, if
α = β, µ = 0 v0 = 0.
(10)
8-2: Loop
A small block with mass m slides in a vertical circle of radius R on the inside of a circular vertical track. There is no friction between the track and the block. At the bottom of the block’s path, the normal force the track exerts on the block has magnitude Nb. What is the magnitude of the normal force (call it Nt) that the track exerts on the block when it is at the top of its path?
Solution:
A block is on the inside of a circular track, moving at speed v. Let the speed at the bottom be vb and that of the top be vt We want expressions for the normal force at top & bottom. Use Newton’s 2nd law.
Top:
XFy = −mg − Nt= −mvt2
R (11)
Nt = mvt2
R − mg (12)
Bottom:
XFy = Nb− mg = mv2b
R (13)
Nb = mg + mv2b
R (14)
Now we need to solve for the velocities. Since we are given Nb and m, we can use equation 14 to solve for vb:
vb2 = R
m(Nb− mg) (15)
Now to find the solutions to the normal force at the top, we need to find vt. Using conservation of energy between the top and the bottom:
KEb+ Ub = KEt+ Ut (16)
1
2mv2b + 0 = 1
2mv2t + 2mgR (17)
vt2 = v2b − 4gR (18)
Now finally we find Nt, using equation 14 one more time:
Nt= mvb2
R − 5 mg = Nb − 6mg . (19)
8-3: Pendulum
Young and Freedman (13th edition) problem 7.82, page 239.
(same as Young and Freedman (12th edition) problem 7.14, page 240.)
Note: Young and Freedman stated this problem with numerical values m = 0.12 kg, ` = 0.80 m, and θ = 45◦. You, however, are being asked to solve the problem symbolically.
A small rock with mass m is fastened to a massless string with length ` to form a pendulum. The pendulum is swinging so as to make a maximum angle of θ with the vertical. Air resistance is negligible.
(a) What is the speed of the rock when the string passes through the vertical position?
Solution:
Choose y = 0 at the lowest point of the swing.
We want to use energy conservation. The tension ~T is perpendicular to the direction of motion, thus does no work; air resistance is negligible; only gravity does work.
E1 = E2, K1+ U1 = K2+ U2 (20)
Since velocity is 0 at the top of the swing, and gravitational potential energy is
U1 = mg`(1 − cos θ), (21)
so
mg`(1 − cos θ) = 1
2mv22 (22)
v2 =p
2g`(1 − cos θ) (23)
(b) What is the tension in the string when it makes an angle of θ with the vertical (i.e., when it is at its maximum angle)?
Solution:
At θ from the vertical, the rock is at the top of the swing, thus has 0 speed.
Apply PF = m~~ a to the radial direction ⇒ T − mg cos θ = 0.
The tension in the spring is
T = mg cos θ (24)
(c) What is the tension in the string as it passes through the vertical?
Solution:
We now consider when it passes through the vertical (point 2). Again, let’s apply Newton’s 2nd law.
XFy = may (25)
⇒ T − mg = mv22
` . (26)
Thus
T = mg + mv22
` (27)
T = mg(3 − 2 cos θ) (28)
The tension increases from point 1 to 2.
8-4: Potential Energy
Young and Freedman (13th edition) problem 7.86, page 239.
(same as Young and Freedman (12th edition) problem 7.86, page 246.) A particle moves along the x-axis while acted on
by a single conservative force parallel to the x-axis.
The force corresponds to the potential-energy func- tion graphed in Fig. P7.86. The particle is released from rest at point A.
(a) What is the direction of the force on the particle when it is at point A?
Figure P7.86 Solution:
F (xA) = − dUdx
x=xA = positive, i.e. rightward.
(b) At point B?
Solution:
F (xB) = − dUdx
x=xB = negative, i.e. leftward.
(c) At what value of x is the kinetic energy of the particle a maximum?
Solution:
E = K + U is fixed, so K is maximum when U is minimized ⇒ x ≈ 0.75 m.
(d) What is the force on the particle when it is at point C?
Solution:
F (xC) = − dUdx
x=xC = 0.
(e) What is the largest value of x reached by the particle during its motion?
Solution:
xmax is the value of x when all energy is potential again, i.e. when a horizontal line through the starting point A intersects blue curve. That is xmax≈ 2.1 m.
(f) What value or values of x correspond to points of stable equilibrium?
Solution:
The minima of U are stable equilibrium:
x ≈ 0.75 m, and (29)
x ≈ 1.8 m. (30)
(g) Of unstable equilibrium?
Solution:
Maxima & saddle points of U are unstable equilibria
x ≈ 1.4 m (31)
8-5: Force of a Baseball Swing
Adapted from Young and Freedman (13th edition) exercise 8.8, page 269.
(a) A baseball has mass mb. If the velocity of a pitched ball has a magnitude of vi and the batted ball’s velocity is vf in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
Solution:
The initial and final momentum (x-component) of the ball are
p1,x = mbvi (32)
p2,x = −mbvf (33)
Apply impulse-momentum theorem, ~J = ~p2 − ~p1, the x-component of impulse equals the change in the x-momentum
Jx = p2x− p1x= −m(vf + vi) (34) Therefore, the magnitude of the change in momentum of the ball is |Jx| = m(vf+ vi) which is equal to the magnitude of the impulse applied to it by the bat.
(b) If the ball has a mass of 0.145 kg, an initial speed of 45.0 m/s, a final speed of 55.0 m/s, and the ball remains in contact with the bat for 2.00 ms (milliseconds), find the magnitude of the average force applied by the bat.
Solution:
The collision time is ∆t = 2.00 ms = 2.00 × 10−3 s. From Jx = (Fav)x∆t,
⇒ (Fav)x = Jx
∆t = −(0.145 kg)(45.0 m/s + 55.0 m/s)
2.00 × 10−3 s = −14.5 N · s
2.00 × 10−3 s = −7250 N (35) Thus the bat applies an average force of 7250 N to the ball in the x-direction.
8-6: Throwing a Rock
Adapted from Young and Freedman (13th edition) exercise 8.28, page 270.
(a) You are standing on a large sheet of frictionless ice and holding a large rock. In order to get off the ice, you throw the rock so it has velocity vr relative to the earth at an angle θ above the horizontal. If your mass is M and the rock’s mass is mr, what is your speed after you throw the rock?
Solution:
Consider the rock and yourself together as one system. Then since there is no horizontal external force acting on this system, the x-component of the momen- tum of the system is conserved,
⇒ px1 = p2x. Since initially you and the rock are at rest, p1x = 0. After you throw away the rock, the total momentum (x-component) is the sum of the mo- mentum of you and the rock.
⇒ p2x = mrvrx+ mpvpx where mr, mp are the masses of the rock and you, re- spectively, and vrx, vpx are the x-components of the velocities.
⇒ vpx= p2x− mrvrx
mp = −mr
mpvrx = −mr
mp vrcos θ . (36)
(b) Calculate for the following values: vr = 12.0 m/s, θ = 35.0◦, M = 70.0 kg, mr = 15.0 kg.
Solution:
vrx = vrcos 35◦ = 12.0 m/s × cos 35◦ = 9.83 m/s (37)
⇒ vpx = −15.0 kg
70.0 kg · 9.83 m/s = −2.11 m/s (38)
(a negative sign means you are moving in the opposite direction of the rock).
Since you are confined to move on the horizontal surface
⇒ ~vp = vpxˆı = −(2.11 m/s) ˆı (39) Your speed after throwing the rock is 2.11 m/s.
8-7: Speed of a Bullet
Adapted from Young and Freedman (13th edition) exercise 8.42, page 271.
(a) A bullet with mass mb is fired horizontally into a wooden target block of mass Mt resting on a horizontal surface. The coefficient of kinetic friction between block and surface is µ. The bullet remains embedded in the block, which is observed to slide a distance d along the surface before stopping. What was the initial speed of the bullet?
Solution:
Set the initial speed of the bullet to be vb. Consider the bullet and the block together as the system. Assume during the first stage, the bullet embeds itself in the block so quickly that the impulse due to the friction between the surface and block (external force) is negligible. Then we can apply momentum conservation to find the velocity of the system after the bullet embeds in the block and they
move together with the same speed v1. By momentum conservation, p1 = p0.
p1 = (mb+ m)v1, p0 = mbvb (40)
⇒ v1 = mbvb
mb+ m (41)
For the second stage, we use the work-energy theorem.
∆K = Wf (42)
(since the potential energy doesn’t change).
Final speed is 0 ⇒ ∆K = −12mv2
Wf = −f d, f = µN = µmg (43)
⇒ −1
2mv12 = −µmgd (44)
⇒ v12 = 2µdg (45)
⇒
mbvb mb+ m
2
= 2µgd (46)
⇒ vb = p
2µgdmb+ m
mb (47)
(b) Calculate using mb = 5 g, Mt = 1.20 kg, µ = 0.20, d = 0.230 m.
Solution:
vb = p
2µgdmb+ m
mb (48)
= q
2 × 0.20 × 9.8 m/s2× 0.23 m5.00 × 10−3 kg + 1.20 kg
5.00 × 10−3 kg (49)
= 229 m/s (50)
Therefore the initial speed of the bullet is 229 m/s.
8-8: The Roller Skater and the Football
BCG problem 5B.4, page 175. This problem has a hint on page 194, an answer to the hint on page 196, and an answer to the problem on page 197.
You are roller-skating peacefully down a street at a constant speed of 3 m/s when someone suddenly throws a football at you from directly ahead. If your mass is 65 kg and the football’s is 0.40 kg, what is your speed afterwards if
(a) you catch the ball, which was thrown at you with a horizontal velocity of 15 m/s;
(b) you miss it and it bounces off you with a velocity of 10 m/s (relative to the street) in the opposite direction?
In each case, what is the total kinetic energy of the system comprising you and the ball before and after your interaction? (Remember: solve the answer in terms of variables before you start sticking in numbers!)
Solution:
In this problem we will use conservation of momentum!
Let’s name our variables first. Let the mass of the person be M , the mass of the football be m. The initial velocity of the person is v and that of the ball vb. The final speed of the person is vf and if the ball is not caught by the person, the speed will be vb,2 but of course if the person catches the ball, the latter will have the same speed vf as the skater.
A good habit is to write down the momenta before and after the collision. For the first case,
quantity Before collision After collision
px M v − mvb (M + m)vf
KE 12M v2+12mvb2 = 337.5 J 12(M + m)vf2
By conservation of momentum, in the first case, the final velocity is:
vf = M v − mvb
M + m = 65 kg × 3 m/s − 0.4 kg × 15 m/s
65 kg + 0.4 kg = 2.89 m/s (51) For the second case:
quantity Before collision After collision
px M v − mvb M vf + mv2
KE 12M v2+12mvb2 = 337.5 J 12M vf2+ 12mv22
vf = M v − m(vb+ v2)
M = 65 kg × 3 m/s − 0.4 kg(15 m/s + 10 m/s)
65 kg = 2.85 m/s (52)
Knowing these values for vf, we can calculate the final kinetic energy for each of these cases:
Case 1: Kf = 1
2(M + m)vf2 = 273.1 J (53)
Case 2: Kf = 1
2M vf2 +1
2mv22 = 283.3 J . (54)
