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Complete solutions to the homework problems are posted in the display cases in the sixth floor center corridor in McCarthy hall. Copies also may be checked out from the Physics Department Office (MH-611).

Chapter 1 Hints:

1-7:  The volume of a semicircular disk is (1/2)πR²h, where R is the radius of the semi-circle and h is the height of the disk.  Using the information given in the book, you can calculate the volume of the semicircular disk in m³.  Now use the fact that 1m³ = (100 cm)×(100 cm)×(100 cm) = 10⁶cm³ to find the number of cubic centimeters of ice in Antarctica.

1-17:  We don't know which, if any, of the three clocks is keeping the correct time.  But we can compare the times read by each clock if we assume that each ticks a fixed rate.  Let's take clock A as our reference.  Then clock B ticks at a rate equal to (290-125)/200 = 0.825 times clock A's rate.  And, when clock A reads zero seconds, clock B reads (125 s - 0.825×312 s) = -132.4 s.  Clock C ticks at a rate equal to (142-92)/175 = 0.2857142 times clock B's rate = 0.2357142 times clock A's. rate.  When clock B reads zero, clock C reads (92s - 0.2857142×25s) = 7.142855 s.  So for part (a) events 600 s apart on clock A will be 600 s × 0.825 = 495 s on clock B.  For part (c) when clock A reads 400 s clock B will read -132.4 s + 400 × 0.825 = 197.6 s.

1-18:  The key to this problem is the word "cumulative". If the length of the day increases by 0.001 s per century, after 20 centuries the length of the day is 0.020 s longer. Thus, the average day during this period is 0.010 s longer than the starting day of the 20 century period. If you use the length of the day as your clock, then compared to a clock that is not slowing down, you have "lost" an amount of time equal to (0.010 s/day)×(365.25 day/year)×(2000 years).

Chapter 2 Hints:

2-4:  The distance covered during each part of the trip is the same, call it d.  By definition the average speed is the total distance traveled divided by the total time the trip took.  The time to travel uphill is d/s₁, while the time to travel downhill is d/s₂, where s₁ and s₂ are the uphill and downhill speeds.  So, from the definition of average speed, the average speed for the whole trip is 2d/(d/s₁+d/s₂) = (s₁×s₂)/(s₁+s₂).  (Note that solving the problem algebraically first is more efficient.)

2-9:  Let L₁ = 1 km be the true length of track 1, and L₂ = L₁+∆L be the true length of track 2.  Then the average speeds of the two runners are s₁ = 1 km/27.95 s and s₂ = (1 km + ∆L)/28.15 s.  For runner 1 to be faster, it must be the case that s₁ > s₂.  Or, 1 km/27.95 s > (1 km + ∆L)/28.15 s.  Use this result to find the limit on ∆L.

2-13:  By definition the instantaneous velocity is dx/dt = 4.5t², where t is in seconds.  Parts (a) through (d) are straight-forward applications of the definitions.  For part (e) you need to know the time at which the particle reaches the midpoint between x(2.00 s) = 21.75 m and x(3.00 s) = 50.25 m, or the time at which the particle is at 36 m.  That time is ((36 m - 9.75 m)/1.5 s³/m)^1/3 = 2.596 s.

2-16:  The instantaneous velocity v = dx/dt =(16e^-t-16te^-t).  Setting this equal to zero gives t = 1 s.  Note that the acceleration does not have to be zero for the instantaneous velocity to be zero.

2-19:  For part (c) note that you can find the maximum by setting dx/dt = 2ct - 3bt² equal to zero and solving for t with the given values of c and b.  For part (d) note that the displacement by definition during the time interval from t = 0.0 s to t = 4.0 s is x(4.0 s) - x(0.0 s).  From part (c) you know that the particle reverses direction at t = 2c/3b = 1.0 s, so the total distance traveled is given by x(1.0 s) - x(0.0 s) + |x(4.0 s) - x(1.0 s)|, where the bars indicate absolute value.

2-34:  Since the train accelerates from rest, the velocity and position during the acceleration phase are given by v = at and x = (0.5)at², where a = 1.34 m/s².  Thus, the train reaches the midpoint between stations (403 m) at t = (806 m/1.34 m/s²)^1/2 = 24.53 s.  The maximum speed then is at = 32.87 m/s.  By symmetry the train must decelerate at a = -1.34 m/s² in order to stop at the next station, so it takes the same amount of time slowing down as it did speeding up, and the travel time between stations is 49.06 s.  Accounting for the 20 s stop, the average speed between stations is 806 m/(49.06 s + 20 s) = 11.67 m/s.

2-37:  We can assume that the passenger train is at x = 0 m and moving at 161 km/hr = 44.722 m/s when the engineer first sees the other locomotive and begins to decelerate.  The equations of motion for the passenger train are: (1) v₁(t) = 44.722 m/s - at and (2) x₁(t) = 0 + (44.722 m/s)t - (0.5)at², where a is the magnitude of the deceleration and t the time in seconds after deceleration begins.  The equations of motion for the other locomotive are (3) v₂(t) = 8.056 m/s and (4) x₂ = 676 m + (8.056 m/s)t, since it is moving at constant speed.  For the passenger train not to hit the other locomotive it must slow down to the same speed as the locomotive at the time it just catches up to the locomotive.  That means that v₁(t) = v₂(t) when x₁(t) = x₂(t).  This gives you two equations in two unknowns, a and t, that can be solved for the unknown deceleration a.

2-47:  The key to this problem (pardon the pun) is to determine the time it takes the key to fall.  The boat, moving at constant velocity, must move 12 m in the same time.

2-51:  The velocity of the ball just before it hits can be found from eq. 2-16:  v² = v₀² + 2a(x - x₀), with v₀ = 0 m/s, a = -9.8 m/s², and x₀ = 4 m.  Thus, the downward velocity of the ball just before it hits is -8.854 m/s.  Eq. 2-16 also can be used to find how fast the ball must be moving just as it leaves the ground for it to reach a height of 2.0 m.  Now v = 0, since the ball stops momentarily when it reaches the height of 2.0 m, a is still -9.8 m/s², x = 2.0 m, and x₀ = 0 m.  This implies that v₀ = +6.261 m/s.  The change in velocity is +6.261 m/s - (-8.854 m/s) = +15.115 m/s.  The average acceleration during the time the ball is in contact with the ground, multiplied by the time it is contact must equal this change in velocity.

2-56:  The equations of motion (vertical component) for the basketball player on his way up are v = v₀ - (9.8 m/s²)t and y(t) = v₀t - (4.9 m/s²)t². Using the information that the player reaches a height of 76.0 cm = 0.76 m and the fact that his velocity is 0 at the top of his travel, you can find v₀.  The equations of motion for the player on his way down are a bit simpler.  v = -(9.8 m/s²)t and y(t) = 0.76 m - (4.9 m/s²)t² (assuming you start measuring time from the top of his travel).  From these equations you can find the times he reaches 0.15 m, 0.61 m and 0.76 m on the way up, and the times he reaches 0.61 m, 0.15 m and 0.0 m on the way down.

Chapter 3 Hints:

3-22:  Take the origin of coordinates at oasis A.  Then take the x-axis in the easterly direction, and the y-axis in the northerly direction.  Let vector a = (25 km)i + (0 km)j represent the displacement between oasis A and oasis B.  Then b = (24 km)cos(-15º)i + (24 km)sin(-15º)j represents the camel's first displacement; and, c = 0i + (8.0 km)j represents the camel's second displacement.  That puts the camel at a location given by b + c from oasis A.  The vector representing the camel's displacement from this point to oasis B is just d = a - (b + c).  Then, find the length of d.

3-36:  For part (d) note that d₁•d₂ = |d₁||d₂|cos(θ), where θ is the angle between the two vectors.  So, since the projection of d₁ on the direction of d₂ is |d₁|cos(θ), it can be found by computing (d₁•d₂)/|d₂|.  Note I have used |a| to mean the length of vector a = (a_x² + a_y² + a_z²)^1/2.

3-37:  For part (a) recognize that the points D, A, C, and B all lie in a plane (the fault plane).  Thus, the Pythagorean Theorem can be used to determine the length of the line AB.  For part (b) recognize that the vertical projection of the line AD is also the vertical projection of the line AB.

3-39:  Let d be the displacement vector from the point P when it is in contact with the floor to the point P when it is at the top of the wheel.  Then |d|cos(θ) = πR, where R is the radius of the wheel.  Likewise, |d|sin(θ) = 2R.  From these two equations you can find both |d| and θ.

3-41:  A + B = 6.0i + 1.0j and A - B = -4.0i + 7.0j.  Adding the two vectors gives 2A = 2.0i +8.0j.  Use that information to find the magnitude of A.

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Chapter 4 Hints:

4-8:  Take the positive x-axis in the easterly direction, and the positive y-axis in the northerly direction.  Then, the plane's total displacement is the sum of two displacement vectors: a = (483 km)i + 0j and b = 0i - (966 km)k.  Use the Pythagorean Theorem to find the magnitude of a + b, and tan(θ) = (a + b)_y/(a + b)_x to find the direction of a + b.  Use the definition of average velocity, Δr/Δt = (a + b)/Δt, to find the average velocity vector.  Finally, note that the average speed is defined as the total distance traveled divided by Δt.

4-16:  For a collision to occur there has to an instant of time t when the positions of both particles are the same; i.e., x_A(t) = x_B(t) and y_A(t) = y_B(t).  So, we have two equations in two unknowns: (3.0 m/s)t = 0.5(0.4 m/s²)t²sin(θ) in the x-direction, and 30 m = 0.5(0.4 m/s²)t²cos(θ) in the y-direction.  From the first equation, we find that t = (15 s)/sin(θ).  Using that value of t in the second equation, we have 30 m = (45 m)cos(θ)/sin²(θ), or sin²(θ) = 1.5cos(θ).  This can be solved either algebraically by substituting 1 - cos²(θ) for sin²(θ) and obtaining the quadratic equation cos²(θ) + 1.5cos(θ) - 1 = 0 and solving for cos(θ) (use the positive root).  An alternative approach would be to plot both sin²(θ) and 1.5cos(θ) over the range 0 to 90 degrees, observing the angle where the plots intersect.

4-28:  Ignoring air resistance, the equations of motion for the position of the soccer ball are y(t) = (19.5 m/s)sin(45 deg)t - (4.9 m/s²)t² and x(t) = (19.5 m/s)cos(45 deg)t.  From the first of these, we find that the soccer ball hits the ground at t = 2.814 s.  It's x-position at that time, from the second equation, is 38.8 m. So the player has to cover 55 m - 38.8 m = 16.2 m in 2.814 s, yielding an average speed of 5.76 m/s.

4-41:  The ball can be treated as a point particle.  Since there is no acceleration in the x-direction v_x is constant until the ball hits a step.  Thus, the ball reaches the edge of the first step at t₁ = (0.203 m)/(1.52 m/s) = 0.1336 s, the edge of the second step at t₂ = (2 × 0.203 m)/(1.52 m/s) = 0.2671 s, etc.  In the y-direction the ball is accelerating at -9.8 m/s², so its y-position is given by y(t) = -(4.9 m/s²)t² taking the top of the stairway to be at y = 0.0 m, and noting that v_y0 = 0 m/s.  So at t₁, y = -0.0874 m, at t₂, y = -0.3496 m, at t₃, y = -0.7866 m, etc.  Now the height of the first step is -0.203 m, the second -0.406 m, the third -0.609 m, so the ball misses the first and second steps but hits the third step.

4-50:  In uniform circular motion the acceleration vector always points toward the center of the circle that the particle is moving around.  At t₁ = 4.00 s the acceleration vector points to the right (positive x-direction) along the line y = 6.00 m.  At t₂ = 10.00 s the acceleration vector points in the positive y-direction.  This implies that the particle has traveled through three-quarters of the circumference of the circle in 6.00 s.  Since the speed of the particle is 3.00 m/s, this implies that (3/4)×(2πr) = (3.00 m/s)×(6.00 s).  From this equation, we can find the radius of the circle r, and the center of the circle must lie at (5.00 m + r, 6.00 m).

4-53:  Once the string breaks the stone moves like any other projectile that starts out 2.0 m above ground with an initial velocity v_x0 in the horizontal direction.  In the vertical direction the equations of motion for the stone are v_y(t) = -gt and y(t) = 2.0 m - (4.9 m/s²)t².  From the second of these equations we can find the time t' it takes for the stone to hit the ground.  We also know that during this time interval the stone has traveled horizontally a distance of 10.0 m.  So v_x0t' = 10.0 m.  But v_x0 is the speed that the rock was moving in its circular path, so the centripetal acceleration must have been (v_x0)²/r, where r was the radius of the circular path the stone was moving along before the string broke.

4-57:  When the man is running in the same direction as the moving sidewalk is moving, his velocity relative to the ground is v + V, where v is the man's speed relative to the moving sidewalk and V is the speed of the moving sidewalk relative to the ground.  When he is running in the opposite direction to the velocity of the moving sidewalk his velocity is -v + V relative to the ground.  Recalling that speed is the absolute value of the velocity, we note that in the first case he takes a time t₁ = 2.50 s = L/(v+V), where L is the length of the moving sidewalk.  Running in the other direction he takes a time t₂ = 10.0 s = L/(v-V).  So (10.0 s)/(v-V) = (2.0 s)/(v+V).  This is enough information to find the ratio v/V.

4-66:  Let the vector representing the velocity of the boat (relative to the water) be v = v_xi + v_yj, taking the x-direction to be parallel to the stream and the y-direction to be perpendicular to the stream.  Note also that (v_x² + v_y²)^1/2 = 4.0 m/s.  The vector representing the velocity of the water (relative to the ground) is V = (1.1 m/s)i.  For the boat to move in a straight line to the clearing it's velocity vector relative to the ground, v + V, must point directly at the clearing.  Let θ be the angle between the perpendicular to the river and the line that points directly at the clearing.  Then tan(θ) = (v_x - 1.1 m/s)/v_y.  Since we know that tan(θ) = (82 m)/(200 m) = 0.41, we have enough information to solve for the v_x and v_y.

Chapter 5 Hints:

5-5:  Since F = ma, the net force vector must be in the same direction as the acceleration vector.  Thus F₁ + F₂ must make an angle of 240 degrees with the x-axis.  In addition, F₁ + F₂ = a/m.  So, F_1x + (20 N) = (0.5 kg^-1)×(12 m/s²)cos(240 deg) = -3 N.  And, F_1y = (0.5 kg^-1)×(12 m/s²)sin(240 deg) = -5.196 N.  So, F₁ = (-3 N)i - (5.196 N)j.

5-27:  Assuming that the girl maintains a steady 5.2 N force on the rope, the acceleration of the sled will be a_sled = -F/m_sled, where the minus sign indicates that the sled accelerates towards the girl.  From Newton's third law, there is a force of equal magnitude (but opposite direction) on the girl, so a_girl = F/m_girl.  Since both the girl and the sled move with constant accelerations, their positions are given by x_girl(t) = (0.5)|a_girl|t² and x_sled(t) = (15.0 m) - (0.5)|a_sled|t².  The girl and the sled meet when their positions are the same, so you can use these two equations to find the time that they meet, then the corresponding value of x.

5-43:  The applied force, 3.2 N, must accelerate both blocks.  So the acceleration a = F₁/(m₁+m₂) = 0.9143 m/s².  The contact force between the two blocks, F₂, is what accelerates the block on the right.  So, F₂ = m₂a = 1.097 N.  Note that if the 3.2 N force is applied to the block on the right, the contact force, F₂^', now has to accelerate the block on the left, which is more massive, so F₂^' must be larger than F₂.

5-50:  For part (a) note that there are two upward forces and one downward force on the combined mass of the man and the bosun's chair.  There is an upward force on the man's hands that from Newton's third law is just equal to the force he pulls with.  There also is an equal force at the end of the rope connected to the bosun's chair.  The downward force is just the weight of the man and the bosun's chair.  So 2×F = mg in this case.  For part (b) the net force on the bosun's chair must equal ma.  So 2F - mg = ma.  For parts (c) and (d) note that there now is only one upward force on the bosun's chair.  For the remaining parts isolate the pulley, then note that there is an upward force on the pulley from the ceiling, and two downward forces from the ropes.

5-54:  As long as the cord connecting the two blocks remains taught, both blocks will accelerate at the same rate, i.e. |a₁| = |a₂| = a.  Let the tension in the cord be T, then for the horizontal block F + T = m₁a and for the block on the incline m₂gsin(30 deg) - T = m₂a.  Note that m₂gsin(30 deg) is the downhill component of the gravitational force on m₂.  Be sure you understand why this so.  Since F and the masses are known, the two equations can be solved for the tension.  For the second part of the problem, note that the condition for the cord to go slack is that F is just large enough to accelerate m₁ at rate that is larger than the maximum downhill acceleration that m₂ can have.

5-63:  Note that the weight of an object is equal to its mass × g.  In other words, it is the force of gravity on the object.  Since the upward component of F is (450 N)sin(38 deg) = 277.0 N, the crate never lifts of the floor.  So, you only need to compute the net horizontal force to determine the acceleration, but note that in part (b) the mass of the crate is different than in part (a).

Chapter 6 Hints:

6-15:  For part (a) note that the normal force on the block will have a magnitude of 12.0 N (from Newton's third law).  So, the maximum frictional force in the upward direction will be f_s,max = μ_sF_N = 0.6×(12.0 N) = 7.2 N.  Since this exceeds the 5.0 N weight of the block, it will not slide provided it was stationary to begin with.  But note that if the block had been sliding to begin with, the upward frictional force is not large enough to balance the weight.

6-18:  The component of the net force on the car along the incline is F_net = mgsin(12 deg) - μ_kmgcos(12 deg) = ma, where m is the mass of the car and a is its acceleration; and, we have taken the downhill direction to be positive.  (You should draw a free body diagram of the forces on the sliding car to make sure that you understand how the equation for F_net was obtained.)  Note that we don't have to know m to find the acceleration.  Once we have found a (which should be a negative number since the car is decelerating), we can use Eq. 2-16 v² = v₀² + 2a(x - x₀) with v₀ =18 m/s and (x - x₀) = 24.0 m to find the velocity of the moving car when it hits the stationary one.

6-23:  Let the tension in left-hand cord be T₁ and the tension in the right-hand cord be T₂.  Then (1) T₁ - Mg = Ma, (2) T₂ - T₁ - μ_k(2Mg) = 2Ma, and (3) T₂ - 3Mg = -3Ma, where a is known and equal to 0.500 m/s².  Use equations (1) and (3) to find T₁ and T₂ in terms of M, g, and a.  Substitute in equation (2) to find μ_k.

6-30:  The maximum frictional force on m₂ = μ_sN₂, where N₂ is the magnitude of the normal force on m₂ = m₂g.  So, f_2,max = 0.6×(10 kg)×(9.8 m/s²) = 58.8 N.  This is less than the force applied to m₂, so the block slides on the slab.  With the block sliding, the net horizontal force on the block is (-F + μ_km₂g)i = m₂a_2,xi.  So a_2,xi = -(6.08 m/s²)i.  The only horizontal force on the slab is the frictional force on it, which from Newton's third law is  -μ_km₂gi = m₁a_1,xi.  So, a_2,xi = -(0.4×(10 kg)×(9.8 m/s²))/(40 kg)i = -(0.98 m/s²)i.

6-46:  If m is the mass of the strap and T is the tension on the strap, then if the streetcar is moving in a straight line at constant speed T would be equal to mg.  But, when the streetcar rounds the curve T must provide the centripetal force and balance the weight of the strap.  So, the horizontal component of T, Tsin(θ) = mv²/R taking θ to be the angle between the strap and the vertical direction.  And, the vertical component of T, Tcos(θ) = mg.  Thus tan(θ) = v²/(Rg).

6-51:  Let T_upper represent the tension in the upper cord, and T_lower the tension in the lower cord.  Let θ be the angle between the two cords.  Then the half-angle θ/2 can be found from sin(θ/2) = (d/2)/L.  But since d = L, sin(θ/2) = 0.5 and θ/2 = 30 deg.  This also can be seen by noting that d and the two cords form an equilateral triangle.  In the vertical direction |T_upper|sin(θ/2) - |T_lower|sin(θ/2) = mg.  Since the magnitude of T_upper  and m are known (35 N and 1.34 kg), this is enough to solve for the magnitude of T_lower.  But the horizontal components of the two tension forces provide the centripetal force.  So,  |T_upper|cos(θ/2) + |T_lower|cos(θ/2) = mv²/Lcos(θ/2), which can be solved for the speed of the ball (part c).  For parts b and d, let T_sum = T_upper + T_lower then find the components of T_sum.

Chapter 7 Hints:

7-5:  Let m be the mass of the father and v_0,f be his original speed.  Let v_0,s be the son's original speed.  Then translating the statement of the problem into algebraic equations, we have (1/2)mv_0,f² = (1/2)²(m/2)v_0,s² initially (note the two factors of 1/2 on the right hand side).  When the father speeds up by 1.0 m/s, we have (1/2)m(v_0,f + 1.0 m/s)² = (1/2)(m/2)v_0,s².  Notice that m cancels out of both equations.  We now have enough information to solve for both initial velocities.

7-13:  Since the work done by a constant force is F·d, where d is the displacement, the net work done by the three forces is (-5.0 N)(-3.00 m) + (9.0 N)cos(60 deg)(-3.00 m) = -1.5 J.

7-23:  During the the first displacement the net force on the block of cheese is (3.0 N) - mg, where m is the mass of the cheese.  So the net force on the cheese is 0.55 N.  From F_net = ma, we find that the cheese (and the elevator) is accelerating upward at 2.2 m/s².  The net force on the elevator plus cheese is in magnitude F_cable - (m + M)g = (m + M)a, where M is the mass of the elevator cab.  So F_cable = (900.25 kg)(9.8 m/s² + 2.2 m/s² ) = 10,803 N.  Thus, the work done by the cable during the first displacement is (10,803 N)(2.40 m) = 25,927.2 J.  For part (b) (F_cable)(10.5 m) = 92,610 J, so F_cable = 8,820 N.  This allows us to compute the new acceleration of the elevator cab (plus cheese) from F_cable - (m + M)g = (m + M)a.  This implies that a = -0.00272 m/s² (the elevator is decelerating).  As before the net force on the cheese is F_N - mg = ma.  So F_N = (0.25 kg)(9.8 m/s² - 0.00272 m/s²) = 2.449 N.

7-28:  From Fig. 7-36 we can infer that the force we apply just balances the spring force as we extend the spring, so the work done by our externally applied force must be W = ∫(kx)dx = (1/2)kx², where k is the spring constant.  From Fig. 7.36 W = 1.0 J when x = 0.03 m, so k = 2222.2 N/m.  For part (a) W_spring = -(1/2)k(x₂² - x₁²), where x₂ is the position at the end of the displacement and x₁ is the position at the start of the displacement.  So W_spring = 10,000 J.

7-38:  From the work-kinetic energy theorem W = ΔK, so ∫(cx - 3.00x²)dx = [(1/2)cx² - x³] evaluated at an upper limit of 3.0 m and a lower limit of 0.0 m gives us the work done; and this has to be equal to the change in K.  So (1/2)c(9.0 m²) - 27 J = -9.0 J, which implies that c = 4 J/m².

7-45:  Since there are small accelerations during the trip we don't know for sure that the speed of the elevator cab is constant (although it probably is for most of the trip).  But we do know that the elevator cab starts at rest and ends at rest, so we can compute the work that F_cable must have done to lift the elevator cab.  Since the kinetic energy of the system has not changed, the work done by F_cable plus the work done by the gravitational force must add to zero.  Thus W_cable - Mgd + mgd = 0, where M is the mass of the elevator cab (note the work done by gravity on the cab is negative because the cab moved up while the force of gravity was down) and m is the mass of the counterweight (note the work done by gravity on the counterweight is positive because the counterweight moves down).  so W_cable = (1200 kg - 950 kg)(9.8 m/s²)(54 m) = 1.323×10⁵ J.  The average power supplied by the cable is P_ave = W_cable/Δt = (1.323×10⁵ J)/(180 s) = 735 J.

7-50:  Since the spring compresses a distance of 0.12 m, the work done by the gravitational force on the block is just mgd = (0.250 kg)(9.8 m/s²)(0.12 m) = 0.294 J (note that the work is positive because the gravitational force is in the same direction as the displacement).  The work done by the spring on the block is -(1/2)kd² = -(0.5)(250 N/m)(0.12 m)² = -1.8 J.  Since all of the energy of the block has been converted into elastic energy of the spring when the block stops, it must be the case that K + 0.294 J = 1.8 J, where K was the kinetic energy of the block just as it hit the spring.  So K = 1.506 J = (1/2)mv², where v is the speed of the block just as it hits.  So v = 3.471 m/s.  If we were to double the speed of the block at the point of contact with the spring, then K is four times as large.  So, (6.024 J) + mgd' = (1/2)kd'² where d' is the new compression distance.  This is a quadratic equation, 0.5kd'² - mgd' - (6.024 J) = 0, which can be solved for d'.

Chapter 8 Hints:

8-8:  For part (a) note that the ball starts at a height of L - Lcos(θ) above the lowest point; and the work done by the gravitational force W_g = -ΔU_g = -(mg(0.0 m) - mgL(1-cos(30 deg)) = 13.13 J.  For (b) note that ΔU_g = - W_g.  For (c) note that U_g = mgy, where y is the vertical distance from the bottom of the trajectory of the ball.

8-18:  Taking the zero of gravitation potential energy at the point where the spring is maximally compressed we have using conservation of mechanical energy that ΔU_g + ΔU_elastic = 0.  (Note that ΔK = 0, since the block starts at rest and stops momentarily when the spring is maximally compressed.)  So -mg(0.4 m + d) + (1/2)kd² = 0, where d is the distance the spring is compressed.  This is a quadratic equation that can be solved for d (take the positive root).

8-23:  For part (a) note that ΔU_g + ΔK = 0 implies that (1/2)mv² = mgL, where v is the velocity of the ball at the low point.  For part (b) the only difference is that the change in gravitational potential energy is now mg(L - 2r), so (1/2)mv² = mg(L - 2r).

8-35:  Consider the cord to be made up of differential elements of length Δl.  The mass of this element is Δm = ρΔl, where ρ is the density of the cord.  After the cord is in the vertical position, the change in gravitational potential energy of that particular element is -(Δm)gy taking the zero of gravitational potential energy at ceiling height.  So ΔU_g = -gyρΔy, in the limit that Δy goes to zero this becomes dU_g = -gyρdy, so the total change in gravitational potential energy is ∫(-gyρ)dy where the definite integral is from the bottom end of the cord to the top.  So ΔU_g = -(1/2)gρl², where l is the length of the cord and ρ = m/l.  So ΔU_g = -(1/2)mgl.

8-38:  To find the equilibrium distance set dU/dr = 0.  Since U = Ar^--12 - Br^-6, dU/dr = -12Ar^-13 + 6Br^-7 = 0, which yields -2A(r_equil)^-6 + B = 0 or r_equil = (2A/B)^-6.  For part (b) take the second derivative of U with respect to r, and show that at r_equil this is positive, so we have a true minimum.  Thus dU/dr will be positive for values of r greater than  r_equil and negative for values less than  r_equil.  Thus, the forces are attractive in both cases.

8-42:  The work done by the applied force is just Fd = 105 J.  Since the frictional force is f = μ_kmg = 23.52 N in this case, part of the work done by the horizontal applied force F goes into thermal energy and part into increasing the kinetic energy of the block.  The total thermal energy generated can be determined from the work done by the frictional force, W_f = fd = 70.56 J.  Since 40 J of thermal energy went into the block, the remainder, 30.56 J, went into heating the floor (actually a little thermal energy may go into heating the air in the vicinity).  Since only 70.56 J of the 105 J supplied by the external force went to thermal energy, the remainder, 34.44 J, must represent the increase in kinetic energy of the block (assuming the floor is entirely horizontal).

8-61:  Let m be the mass of the block.  The K of the block is (1/2)m(8.0 m/s)² - m(9.8 m/s²)(2.0 m) = m(12.4 m²/s²) at the point it hits the patch with friction.  The frictional force f = μ_kmgcos(30 deg) = m(3.395 m/s²) and the work done by the frictional force (if the block makes it all the way past the patch with friction) is fL = m(2.546 m²/s²).  Thus the block has more than enough energy to make it past the frictional patch.  When it reaches point B its new K = m(12.4 m²/s²) - mg(0.75 m)×sin(30 deg) - m(2.546 m²/s²) = (1/2)mv_B².  Note that m drops out, and you can solve for v_B, the speed at point B.

9-8:  By symmetry the center of mass of the can, itself, is at x_CM,can = 0.06 m, so we can regard all of the mass of the can to be concentrated at that point.  Let's now suppose that the soda fills the can to a height x, then the center of mass of the soda, itself, is at x/2.  The center of mass of the combination then must be at (mx_CM,can + M(x)×(x/2))/(m + M(x)), where m is the mass of the can and M(x) is the mass of the soda remaining in the can.  When the height of the soda is x, M(x) = (x/(0.12 m))×(1.31 kg).  So x_CM = ((0.140 kg)(0.06 m) + (1.31 kg)(x/(0.12 m)(x/2))/(0.140 kg + (1.31 kg)(x/(0.12 m)) = (8.4×10^-3 + 5.458x²)/(0.140  + 10.92x) with x in meters.  To find the minimum we differentiate and set the result equal to zero.  This leads to the following quadratic equation: 59.58x_min²  + 1.620x_min - 0.09052 = 0 with x_min in meters.  The minimum is at about 0.027 m (2.7 cm).  So as the soda drains out of the can the CM drops to this point then rises back up to 6 cm.

9-16:  Since there are no external force on the canoe, its com cannot change when the two passengers exchange seats.  Let's assume that Ricardo is sitting on the right and Carmelita on the left.  Let's measure distances from Carmelita's original position.  So to start with the com is at x_com = ((30 kg)(1.5 m) + (80 kg)(3.0 m))/(m + 110 kg), where m is Carmelita's mass.  After they switch positions x_com = ((80 kg)(0.4 m) + (30 kg)(1.9 m) + m(3.4 m))/(m + 110 kg).  But since the position of the com can't change in this case, we can obtain an equation for m.

9-22:  p_initial = (0.3 kg)(-15 m/s)cos(35 deg)i + (0.3 kg)(-15 m/s)sin(35 deg)j = (-3.69 kg m/s)i + (-2.58 kg m/s)j.  p_final = (0.3 kg)(20 m/s)i + (0.3 kg)(-20 m/s)j = (6 kg m/s)i + (-6 kg m/s)j taking the positive x-direction toward the pitcher and the positive y-direction up.  So Δp = p_final - p_initial = (9.69 kg m/s)i + (-3.42 kg m/s)j.  The magnitude of Δp then can be found from the Pythagorean theorem.

9-30:  The impulse on the lizard from the water (owing to Newton's third law) will be J = F_aveΔt, provided we can determine the average force.  If we consider the lizard's steps to be a series of impacts on the water, we can use F_ave = -(Δm/Δt)Δv, where Δm is the mass of the lizards foot, Δt is the time for a single step, and Δv is  the speed of  the foot as it hits the water (we are assuming that since the lizard pushes down on the water with his foot that it doesn't bounce back).  So F_ave = -(0.003 kg)(-1.50 m/s)/(0.600 s) = 0.0075 N.  Thus J = F_aveΔt = 0.0045 N·sec is the upward impulse from the water.  The downward impulse from the weight of the lizard is -mgΔt = (0.090 kg)(9.8 m/s²)(0.600 s) = 0.529 N·sec.  Since the downward impulse from gravity is more than an order of magnitude larger than the impulse from the slap, the push of the lizards foot on the air bubble provides the primary support for the lizard.

9-44:  Since the total momentum of the system before the explosion was zero, it must remain zero after the explosion.  This implies that particles A and B must fly apart in opposite directions along a straight line.  And, since the energy stored in the compression of the spring is transferred completely to particles A and B in the explosion, their total kinetic energy must be 60 J.  We can express these two statements in two equations:  2mv₁ + mv₂ = 0 and (1/2)(2m)v₁² + (1/2)mv₂² = 60 J, where we have made use of the fact that the mass of particle A is twice that of particle B.  From the first of these two equations, we find that v₁ = -v₂/2.  Inserting that information in the second equation, we have m(v₂²/4) + (1/2)mv₂² = 60 J.  This can be written as (3/2)(1/2)mv₂² = 60 J so K_B = (2/3)(60 J).

9-52:  Even though energy is lost in the collision, momentum is conserved during the instant the collision takes place.  So the momentum immediately before the collision must equal the momentum just after the collision. Thus (3.0 kg)(20 m/s) + (2.0 kg)(-12 m/s) = (5 kg)(v_after) and v_after = 7.2 m/s.  So just after the collision K = (1/2)(5 kg)(7.2 m/s)².  The combined mass will continue to rise until all of this kinetic energy has been converted to gravitational potential energy.

9-63:  Since both the collision of the basketball with the floor and the collision of the basket ball with the baseball are elastic, both momentum and energy are conserved in both collisions.  We can use (1/2)mv² = mgh to determine the velocities of both balls just before the collision with the floor, since m drops out of that equation.  I.e., v = ((9.8 m/s²)(1.8 m))^1/2 = 4.2 m/s.  Since the collision with the floor is elastic, the basketball is moving upward with this speed, and the baseball is moving downward with this speed as they collide.  So momentum conservation requires that Mv - mv = (M - m)v = mv_final and energy conservation requires that (1/2)(M + m)v² = (1/2)mv_final² if we assume that the basketball stops when it collides with the baseball.  Solve for v in the first of these equations, then substitute in the second.  You will find that m = (1/3)M = 0.21 kg.

Chapter 10 Hints:

10-6:  ω = dθ/dt = (8.0 rad/s²)t + (6.0 rad/s³)t² and α = dω/dt = (8.0 rad/s²) + (12.0 rad/s³)t.

10-16:  Because the disk is rotating with constant angular acceleration α, we know that ω(t) = ω₀ + αt and θ(t) = θ₀ + ω₀t + (1/2)αt².  Let's start our clock at the time the disk is rotation at 10 rev/s = 62.832 rad/s..  That simplifies the equations:  ω(t) = (10 rev/s) + αt and θ(t) = θ₀ + (10 rev/s)t + (1/2)αt².  Let t' be the time at which it has both traveled through 60 more revolutions and the time it reaches an angular speed of 15 rev/s.  so (5 rev/s) = αt' and (60 rev) = (10 rev/s)t' + (1/2)αt'².  So substituting for t' in the second equation we have (60 rev) = (10 rev/s)(5 rev/s)/α +(0.5)α(25 rev²/s²)/α² = (62.5 rev²/s²)/α.  So α = 1.042 rev/s².  For part (b) recall that ω = 15 rev/s after the 60 revolutions have been completed, so (15 rev/s) = (10 rev/s) + (1.042 rev/s²)t'.  For part (c) we can ask at what time in the past the angular speed was zero.  So (0 rev/s) = (10 rev/s) + (1.042 rev/s²)t'', where the magnitude of t'' tells us how long it took to reach 10 rev/s.  For part (d) use t'' in the equation for θ(t) (note t'' is a negative number) to find θ₀.

10-28:  Since the linear speeds of the rims of both wheels are the same, when wheel C reaches an angular speed of 100 rev/min wheel A will have an angular speed ω_A = (25/10)ω_C = 250 rev/min since r_Aω_A = r_Cω_C.  So from ω_A(t) = ω_A(0) + αt = αt.  So, t = (26.18 rad/s)/(1.6 rad/s²) = 16.36 s.  (Note we had to convert the angular speed of wheel A to rad/s since the angular acceleration was given in rad/s².)

10-40:  From Table 10-2 (c) we find that I = (1/2)mR² for a disk of mass m and radius R rotating about its central axis.  From the parallel axis theorem I for a disk rotating about an axis parallel to the central axis is I_com + mh² = (1/2)mR² + mh², where h is the distance from the central axis to the parallel axis.  Note that in Fig. 10-36 there are 15 disks.  The mass of each disk then is M/15 = 6.667 mg, and the radius of each disk is L/30 = 0.0333 m.  Thus, the total rotational inertia will be I_tot = (1/2)mR² + 2{[(1/2)mR² + m(2R)²] + ........ + [(1/2)mR² + m(14R)²]}.

10-55:  Let T₁ be the tension in the cord on the left, and T₂ be the tension in the cord on the right.  Then for the block on the left T₁ - mg = ma, for the block on the right T₂ - Mg = -Ma, and for the pulley T₂r - T₁r = Iα = Ia/r.  We also know that -0.75 m = -(1/2)a(5.0 s)² since the block is falling with constant acceleration.  This allows us to solve for a; and, since r, m and M also are known we have enough information to find T₁, T₂, and I from the three equations.

10-64:  For part (a) use the parallel axis theorem to find I.  For part (b) note that the center of mass  of the cylinder has dropped a vertical distance h = 0.05 m when the cylinder reaches its lowest point.  Thus, it has lost gravitational potential energy equal to mgh.  This must be balanced by a corresponding gain in rotational kinetic energy, so at the lowest point mgh = (1/2)Iω² since the cylinder started at rest.

Chapter 11 Hints:

11-9:  As the ball rolls downhill it loses gravitational potential energy and gains rotational and translational kinetic energy.  Thus mg(H - h) = (1/2)I_comω² + (1/2)mv_com².  For a solid ball I_com = (2/5)mR².  So mg(H - h) = (1/2)(2/5)mR²ω² + (1/2)mv_com².  But, Rω = v_com, so mg(H - h) = (1/2)(2/5)mv_com² + (1/2)mv_com² = 0.7mv_com².  This allows us to solve for v_com, which is the horizontal velocity of the ball as it leaves the ramp.  Once it leaves the ramp the center of mass of the ball will fall just like a point particle of mass m, so the rest is straightforward.

11-13:  From conservation of energy Mgh = (1/2)I_comω² + (1/2)Mv_com² = (1/2)βMR²ω² + (1/2)Mv_com² = (1/2)(β + 1)Mv_com², so v_com² = 2gh/(β + 1).  But at the bottom of the track the ball is in uniform circular motion momentarily, so F_N = 2.0Mg = Mg + Mv_com²/r, where r is the radius of curvature of the track at the bottom.  Thus, 2.0g = g + 2gh/[r(β + 1)].  Since h and r are known, we can solve for β.

11-22:  τ = r₁×F, where r₁ = (0.5 m)j - (2.0 m)k, and F = (2.0 N)i - (3.0 N)k.  So τ =