1.Question: (7 points, suggested time 13 minutes)
A stunt cyclist builds a ramp that will allow the cyclist to coast down the ramp and jump over several parked cars, as shown above. To test the ramp, the cyclist starts from rest at the top of the ramp, then leaves the ramp, jumps over six cars, and lands on a second ramp.
H0 is the vertical distance between the top of the first ramp and the launch point.
q0 is the angle of the ramp at the launch point from the horizontal.
X0 is the horizontal distance traveled while the cyclist and bicycle are in the air.
m0 is the combined mass of the stunt cyclist and bicycle.
(a) Derive an expression for the distance X0 in terms of H0, q0, m0, and physical constants, as appropriate.
(b) If the vertical distance between the top of the first ramp and the launch point were 2H0 instead of H0, with no other changes to the first ramp, what is the maximum number of cars that the stunt cyclist could jump over? Justify your answer, using the expression you derived in part (a).
(c) On the axes below, sketch a graph of the vertical component of the stunt cyclist’s velocity as a function of time from immediately after the cyclist leaves the ramp to immediately before the cyclist lands on the second ramp. On the vertical axis, clearly indicate the initial and final vertical velocity components in terms of H0, q0, m0, and physical constants, as appropriate. Take the positive direction to be upward.
Answer/Explanation
Ans:
(a)
\(\sum E_{i} =\sum E_{f}\)
\(m_{0}gH_{0} = \frac{1}{2}m_{0}{v_{0}}^{2}\)
\({v_{0}}^{2} = 2gH_{0}\)
\(v_{0} = \sqrt{2gH_{0}}\)
\(vertical x_{0} = \frac{1}{2}at^{2} + v_{0}sin\theta _{0}t\)
\(o = \frac{1}{2}at + v_{0}sin\theta \)
\(o = \frac{1}{2}(-9.8)t + v_{0}sin\theta \)
\(t = \frac{v_{0}sin\theta}{4.9} = \frac{\sqrt{2gH_{0}sin\theta }}{4.9}\)
\(horizontal: x = vt\)
\(x_{0} = \sqrt{2gH_{0}}. \frac{\sqrt{2gH_{0}}sin\theta }{4.9}\)
\(x_{0} =\frac{2gH_{0}sin\theta _{0}}{4.9}\)
(b)
The cyclist would be able to jump over 12 cars. In the expression in part (a), X0 × H0, so if H0 doubles, then X0 doubles as well. Double the amount of 6 cars (X0) is 12 Cars.
(c)
Question (12 points, suggested time 25 minutes)
A group of students is investigating how the thickness of a plastic rod affects the maximum force Fmax with which the rod can be pulled without breaking. Two students are discussing models to represent how Fmax depends on rod thickness.
Student A claims that Fmax is directly proportional to the radius of the rod.
Student B claims that Fmax is directly proportional to the cross-sectional area of the rod—the area of the base of the cylinder, shaded gray in the figure above.
(a) The students have a collection of many rods of the same material. The rods are all the same length but come in a range of six different thicknesses. Design an experimental procedure to determine which student’s model, if either, correctly represents how Fmax depends on rod thickness. In the table below, list the quantities that would be measured in your experiment. Define a symbol to represent each quantity, and also list the equipment that would be used to measure each quantity. You do not need to fill in every row. If you need additional rows, you may add them to the space just below the table.
Describe the overall procedure to be used, referring to the table. Provide enough detail so that another student could replicate the experiment, including any steps necessary to reduce experimental uncertainty. As needed, use the symbols defined in the table and/or include a simple diagram of the setup.
(b) For a rod of radius r0, it is determined that Fmax is F0, as indicated by the dot on the grid below. On the grid, draw and label graphs corresponding to the two students’ models of the dependence of Fmax on rod radius. Clearly label each graph “A” or “B,” corresponding to the appropriate model.
The table below shows results of measurements taken by another group of students for rods of different thicknesses.
(c) On the grid below, plot the data points from the table. Clearly scale and label all axes, including units. Draw either a straight line or a curve that best represents the data.
(d) Which student’s model is more closely represented by the evidence shown in the graph you drew in part (c) ?
____ Student A’s model: Fmax is directly proportional to the radius of the rod.
____ Student B’s model: Fmax is directly proportional to the cross-sectional area of the rod. Explain your reasoning.
Answer/Explanation
Ans:
(a)
1) Students would measure radius of rod and calevlate the cross section from this.
2) Students would connect force sensor to one end of rod and pull until rod breaks. Record Nos. Repeat with rod of same thicknes,
3) Repeat steps I & Q with Rods of different material to reduce experimental uncertainty.
4) Students will then graph average N and r to Find the line of best Fit and analyze.
(b)
(c)
(d)
Student B claims Fmax is directly proportional r2. Fmax & r2 ‘s relationship would not be linear as shown in the graph.
Question (12 points, suggested time 25 minutes)
(a) A student of mass MS, standing on a smooth surface, uses a stick to push a disk of mass MD. The student exerts a constant horizontal force of magnitude FH over the time interval from time t = 0 to t = tf while pushing the disk. Assume there is negligible friction between the disk and the surface.
i. Assuming the disk begins at rest, determine an expression for the final speed vD of the disk relative to the surface. Express your answer in terms of FH, tf , MS, MD, and physical constants, as appropriate.
ii. Assume there is negligible friction between the student’s shoes and the surface. After time tf , the student slides with speed vS. Derive an equation for the ratio vD / vS. Express your answer in terms of MS, MD, and physical constants, as appropriate.
(b) Assume that the student’s mass is greater than that of the disk (MS > MD). On the grid below, sketch graphs of the speeds of both the student and the disk as functions of time t between t = 0 and t = 2tf . Assume that neither the disk nor the student collides with anything after t = tf . On the vertical axis, label vD and vS. Label the graphs “S” and “D” for the student and the disk, respectively.
(c) The disk is now moving at a constant speed v1 on the surface toward a block of mass MB, which is at rest on the surface, as shown above. The disk and block collide head-on and stick together, and the center of mass of the disk-block system moves with speed vcm.
i. Suppose the mass of the disk is much greater than the mass of the block. Estimate the velocity of the center of mass of the disk-block system. Explain how you arrived at your prediction without deriving it mathematically.
ii. Suppose the mass of the disk is much less than the mass of the block. Estimate the velocity of the center of mass of the disk-block system. Explain how you arrived at your prediction without
deriving it mathematically.
iii. Now suppose that neither object’s mass is much greater than the other but that they are not necessarily equal. Derive an equation for vcm. Express your answer in terms of v1, MD, MB, and
physical constants, as appropriate.
iv. Consider the scenario from part (c)(i), where the mass of the disk was much greater than the mass of the block. Does your equation for vcm from part (c)(iii) agree with your reasoning from part (c)(i) ?
____ Yes ____ No
Explain your reasoning by addressing why, according to your equation, vcm becomes (or approaches) a certain value when MD is much greater than MB.
Answer/Explanation
Ans:
(a) (i)
\(a = \frac{V_{p}}{T_{f}}\)
FH = MS a
\(F_{H} = (M_{D})\left ( \frac{Vp}{Tf} \right )\)
\(F_{H} T_{F} = (M_{D})V_{P}\)
\(V_{D} = \frac{F_{H}T_{F}}{M_{D}}\)
(ii)
\(\frac{V_{D}M_{P}}{V_{S}M_{S}}\)
(b)
(c) (i)
The velocity of the system will be slightly less than V1 . COP – mv = (m + mb)v. Because the mass of the disk-block system is slightly greater than the disk.
(ii)
The velocity of the disk-block system will be much less than V1. Ove to the law of conservation of momentum a larger mass collision decreases velocity.
(iii)
V1 M0 = Vcm (MD + MB)
Vcm = \(\frac{V_{1}M_{D}}{\left ( M_{D} + M_{B}\right )}\)
(iv)
___X___ Yes _______No
Because Vcm is equal to the ratio of MD to MD + MB multiqlied by V1. In this scenario the ratio of MD / (MD + MB) is slightly less than I which would make V1 slightly less than Vcm
Question: (7 points, suggested time 13 minutes)
A cylinder of mass m0 is placed at the top of an incline of length L0 and height H0, as shown above, and released from rest. The cylinder rolls without slipping down the incline and then continues rolling along a horizontal surface.
(a) On the grid below, sketch a graph that represents the total kinetic energy of the cylinder as a function of the distance traveled by the cylinder as it rolls down the incline and continues to roll across the horizontal surface.
The cylinder is again placed at the top of the incline. A block, also of mass m0, is placed at the top of a separate rough incline of length L0 and height H0, as shown above. When the cylinder and block are released at the same instant, the cylinder begins to roll without slipping while the block begins to accelerate uniformly. The cylinder and the block reach the bottoms of their respective inclines with the same translational speed.
(b) In terms of energy, explain why the two objects reach the bottom of their respective inclines with the same translational speed. Provide your answer in a clear, coherent paragraph-length response that may also contain figures and/or equations.
Answer/Explanation
Ans:
(a)
Ug ↓ Ke↑ , until h = 0, then
ke = constant
cm = center of mass
(b)
Cylinder : Ug → KEtran. + KErot
block : Ug + Wfriction = KEtran
Ei + Wext = Ef
- The cylinder gains kinetic energy since it’s CM is moving down the incline but Since it is rolling without slipping (ie. not sliding). Some of the initial Ug gets converted into rotational kinetic energy along with translational kinetic energy.
- The block starts at the same height (ie same Ug), but in this case also all of the Ug cant be converted directly into kinetic energy. The Ug gets mostly converted into KE, but some is lost as thermal energy through friction.
- Since the cylinder has energy going into rotational kinetic energy. the block has energy going into thermal energy, both objects end up with a resultant translational kinetic energy, and therefore end with the same speed.
Question: (7 points, suggested time 13 minutes)
Two pulleys with different radii are attached to each other so that they rotate together about a horizontal axle through their common center. There is negligible friction in the axle. Object 1 hangs from a light string wrapped around the larger pulley, while object 2 hangs from another light string wrapped around the smaller pulley, as shown in the figure above.
m0 is the mass of object 1.
1.5m0 is the mass of object 2.
r0 is the radius of the smaller pulley.
2r0 is the radius of the larger pulley.
(a) At time t = 0, the pulleys are released from rest and the objects begin to accelerate.
i. Derive an expression for the magnitude of the net torque exerted on the objects-pulleys system about the axle after the pulleys are released. Express your answer in terms of m0, r0, and physical constants, as appropriate.
ii. Object 1 accelerates downward after the pulleys are released. Briefly explain why.
(b) At a later time t = tC, the string of object 1 is cut while the objects are still moving and the pulley is still rotating. Immediately after the string is cut, how do the directions of the angular velocity and angular acceleration of the pulley compare to each other?
________ Same direction __________ Opposite directions
Briefly explain your reasoning.
(c) On the axes below, sketch a graph of the angular velocity ω of the system consisting of the two pulleys as a function of time t. Include the entire time interval shown. The pulleys are released at t = 0, and the string is cut at t = tC.
Answer/Explanation
Ans:
(a) (i)
τ = Fx.r \(\sum \imath = \imath _{Fg_{1}} – \imath _{Fg_{2}}\)
\(\sum \imath = m_{0}g \left ( 2r_{0} \right ) – 1.5 m_{0}g(r_{0})\)
\(\sum \imath = 2m_{0}gr_{0} – 1.5 m_{0}gr_{0}\)
\(\sum \imath = \frac{1}{2} m_{0}gr_{0}\)
(ii)
Object 1 has a greater radius than object 2, and while object 2 has a greater mass, object 1’s radius is greater than object 2’s mass, so object 1 has a greater torque, so it accelerates downward.
(b)
After the string is cut the pulley will begin to rotate it the opposite direction towards object 2 because of it’s mass, so both the angular velocity and acceleration would go in the same direction.
(c)