Download Assignment 6 solutions

Assignment 6 solutions 1)
You are traveling on a hilly road. At a particular spot, when your car is perfectly horizontal, the
road follows a circular arc of some unknown radius. Your speedometer reads 75.0 km/h, and
your apparent weight is 30% larger than usual. Use g = 9.80 m/s2 for this problem.
(a) Are you at the bottom of a hill or the top of a hill at that instant?
Since we are going in a circular arc there must be one component in the net force we are experiencing which points towards the center of the circular arc and whose magnitude is . We do not care about the net force on the horizontal here, because the car is ``perfectly horizontal’’ and therefore the center of the arc is either above or underneath us, so has to be vertical, directed either up or down. So no horizontal force contributes to . What vertical forces do we have? We have gravity, pulling with down, and one normal force from the ground, , pushing us up. The apparent weight is the actually the amount by which you feel to be pushed by the ground. Remember for instance that in a elevator when the elevator accelerates upward you feel a bit heavier and pushed to the ground, while when it accelerates downward you feel lighter. Therefore the 30% heavier weight is actually just the magnitude of . Since we feel heavier we may compare this to upward acceleration of the elevator. This means that we have some extra acceleration upward. This extra part must be the , therefore the center of the arc must be above our head so we’re at the bottom of the hill. The bottom of a hill.
(b) What is the radius of the circular arc?
Use the fact that net force in the vertical is and call `up’ positive. The magnitude of is as discussed 100%+30% of our weight. Also remember to convert the velocity to m/s, otherwise the units won’t be right. So: 2)
Just before hitting the boards of a hockey rink, a puck is sliding along the ice at a constant
velocity. As shown in the figure, the components of this velocity are 3 m/s in the direction
perpendicular to the boards and 4 m/s parallel to the boards. Immediately after bouncing off the
boards, the puck’s velocity component parallel to the boards is unchanged at 4 m/s, and its
velocity component perpendicular to the boards is 1 m/s in case A, 2 m/s in case B, and 3 m/s in
case C.
(a) Without doing any calculations, rank the three cases based on the magnitude of the impulse
the puck experienced because of its collision with the boards, from largest to smallest (e.g.,
B>A=C).
The magnitude of the impulse is the magnitude of the change in linear momentum. In all three cases the direction of the momentum is reversed. So all three cases are like stopping the puck and then throwing it back with some velocity. So the faster you throw it back, the larger is the momentum you gave to it and thus the larger the impulse. Case C is going with the largest velocity back, therefore has the largest impulse, and then come B and A respectively, so: C>B>A (b) If the puck’s mass is 160 g, find the magnitude of the impulse applied by the boards in case
A.
Remember that impulse and momentum are vectors and the direction of the initial and final velocity matters in the impulse. Take the positive direction to be to the right in the horizontal: 3)
A hockey puck is sliding east at a constant velocity v over some ice. A net force F is then
applied to the puck for 5 seconds.
In case A, the net force is directed west.
In case B, the net force is directed south.
In case C, the net force is directed east.
The magnitude of the applied force is the same in each case. (a) Rank these cases, from largest to smallest, based on the magnitude of the change in
momentum experienced by the puck. Use only "greater than" signs and "equals" signs in your
ranking, such as B>A=C
The magnitude of the change in momentum is: Therefore it doesn’t depend on the direction in which the force was applied. So the magnitude of the impulse is the same for all three cases, only its direction differs. So A=B=C A=B=C -or- A=C=B -or- B=A=C -or- B=C=A -or- C=A=B -or- C=B=A
(b) Rank these cases, from largest to smallest, based on the magnitude of the puck’s final
momentum. Use only "greater than" signs and "equals" signs in your ranking, such as B>A=C
In calculating the final momentum we need the direction of the impulse too, since both the momentum and the impulse are vectors. If we draw the a diagram for the three cases and use tip‐to‐tail method we see that (the bold arrows are the three impulses for the three cases, the faint dotted blue arrow is the initial momentum, and the dashed red arrows are the final ones): C
B
A
C>B>A
4)
A car traveling 50 km/h can be brought to a stop in a particular distance under controlled braking
conditions. For this problem, ignore the reaction time of the driver and find the stopping distance
and stopping time after the brakes are applied.
(a) Assuming the force used to bring the car to rest is the same, how much distance is required to
bring the car to a stop if the car is traveling 100 km/h, twice as fast as it was originally? The
distance in this case is larger than the original distance by a factor of:
This can be done using a few different methods. The most familiar one is by saying that the force is the same and therefore is the same in both cases and the distance is given from: times the previous. Another way to Since the velocity is twice as big this time the distance will be see this is to use mechanical energy equation: Change in mechanical energy=work done on the object.
Which give the same equation if we divide by m and multiply by 2. So the distance is 4 times the previous.
(b) How do the stopping times compare? The stopping time in the second case is larger than the
original stopping time by a factor of:
We may use the impulse equation for this: . In the second case the velocity is twice as big, so since is the same the time will become twice as big. So a factor of 2. 5)
A cart that can move along a straight track has a mass of 2.00 kg and an initial velocity of 4.00
m/s in the negative x-direction. The cart is then subjected to a net force that is initially in the
positive x-direction, but which then eventually switches direction to the negative x-direction, as
shown in the graph of force as a function of time below. The force is directed only in the +x or -x
direction.
(a) Complete the table below to show the cart's momentum and velocity at the indicated times.
Use + and - signs, as appropriate.
The initial momentum is just the given initial velocity times mass: For the momentum at other instances we use the equation for change in momentum: Which is the area under the curve of vs . Since the impulse and force are vectors the sign (direction) matters and the areas above the ‐axis are positive while the ones below it are negative. For example to we have two full squares from 0s to 5s, a triangle, which has area of one half of from two squares, from 5s to 10s above, and the same triangle of area below the axis from 10s to 20s. So we have 2squares+1square‐1square = 2squares. Each square has area . Therefore the change in momentum is in the positive x direction. The initial momentum was ‐
8kgm/s and was going in negative x. So the final momentum is: For 30s we have an additional 2 squares under the axis so that the total change in momentum is zero (we lose the 2squares). Therefore =
Time
Momentum
t=0s
p = -8 kg m/s
Velocity
v = -4m/s
t = 20 s p = 92 kg m/s v = 46m/s
t = 30 s p = -8kg m/s
v = -4m/s
(b) The cart's momentum at t = 20 s is exactly the same as its momentum at t =
At 20s we had a total of +2squares change in momentum. The same thing is true for t=5s. So at t=5s the momentum is the same as t=20s. 6)
A system consists of three balls at different locations on the x-axis. Ball 1 has a mass of 7.00 kg
and is located at x = +3 m; ball 2 has a mass of 4.00 kg and is located at x = -1 m; ball 3 has an
unknown mass and is located at x = -4 m.
(a) If the center of mass of this system is located at x = -2 m, what is the mass of ball 3?
call the masses . The formula for center of mass is plug in the information and solve for : (b) Let’s say that you can make ball 3 as light or as heavy as you like. By adjusting the mass of
ball 3, what range of positions on the x-axis can the center of mass of this system occupy?
Let’s draw a diagram to help us with our reasoning: The center of mass is always somewhere in between the masses, it will never be outside the outermost masses. Intuitively, if the masses were on a rod you expect to be able to balance the rod on some point between the masses only and not on the edges. If one mass is much heavier than the others, the center of mass will be very close to that large mass. Here if mass 3 is many orders of magnitude larger than the other two masses, you would effectively feel that all the mass is just mass 3 and that all mass is located at the position of 3. Since mass 3 is the one with the most negative position (‐4m) its location is the most negative number you can get for the center of mass by changing mass 3. So: The most negative position possible is x =-4m.
The other extreme is when mass 3 is so light compared to the other two that it becomes negligible. In that case you are effectively dealing only with masses one and 2 and the center of mass will be the center of mass of those two masses, namely at 1.55m. The calculation is as follows: The most positive position possible is x =1.55m.
7)
A uniform sheet of plywood measuring 4L by 4L is centered on the origin, as shown in the
figure, where L = 1.80 m. One quarter of the sheet (the part in the first quadrant) is removed.
Where is the center of mass of the remaining piece?
We may use the explanation about the uniformity of the sheet and regard it as four equal square blocks of plywood. At first when all four are there the center of mass is in the middle of all of them, right at the origin. When we have a few parts to our object, we can use the center of mass equation by assuming that all of the mass of each of the parts is sitting at the center of mass of that part. The center of mass of each small square is right in its middle point. The equation for center of mass is Where is the position of the center of mass of each part and is the mass of that part. The mass of each of the small squares is just 1/4 the total mass of the sheet so all are the same. What are the position vectors for the center of mass of each of the 3 blocks? They are as follows: Now plug this information in the equation (remember that when we add vectors, coefficients of and are summed separately and the ‘s do not mix with ‘s). As can be seen from the diagram, and cancel out and we get: So since L = 1.80 m: The center of mass of the remaining piece is located at
x = - 0.6m and y = - 0.6m