As I was studying for this test I came across some things giving me trouble and forgot all about submitting these for help. So, if you can take a look at them and help me see what I'm missing that would be great.
1. A ball has a horizontal range of 21 meters when it is projected horizontally from an altitude of 17 meters. What will be its range if it is projected at an angle of 8 degrees horizontal with the same initial speed? Approximately, how much does its horizontal range change per degree from horizontal?
I really can’t draw the picture that I have in front of me, but I feel it accurately depicts the problem above. I have values of a=9.8m/s^2
H=17m
`ds=21m
I am unclear on how to solve this problem as well as applying the angle of 8 degrees below horizontal.
First solve the problem.
If the ball is projected in the horizontal direction, then the initial vertical velocity is zero. So for vertical motion you know v0, a and `ds. You can therefore find `dt.
The horizontal motion is at constant velocity. The time interval is the same as for vertical motion, so you know the horizontal displacement and the time interval. Use these to find the average horizontal velocity.
Having found that horizontal velocity, now consider an equal initial speed but at angle 8 degrees.
What are the vertical and horizontal components of the initial velocity?
Using the initial vertical velocity (which is no longer zero) you again know v0, a and `ds. Solve for `dt.
Then multiply this by the initial horizontal velocity to get the horizontal displacement.
2. A cart of mass 2kg coasts 95cm down an incline at 3 degrees horizontal. Assume that frictional and other non gravitational forces parallel to the incline are negligible:
What is the component of the cart’s weight parallel to the incline?
Wt component parallel=2kg*9.8m/s^2=19.6N
19.6 N is the force of gravity on the cart, which is not parallel to the incline.
You need to sketch this with the 19.6 N weight vector vertically downward, the x axis parallel to the incline, the y axis perpendicular to the x axis. THe weight vector will be at either 267 deg or 273 deg from the positive x axis. Use the sine and cosine to find the components of the weight vector. The x component will be the force parallel to the incline.
How much work does this force do as the cart rolls down the incline?
19.6N*.95m=18.6J
The 19.6 N force of gravity is not in the direction of the .95 m displacement.
The x component of the gravitational force is in the direction of that displacement.
Using the definition of kinetic energy, determine the velocity of the cart after coasting the 95cm, assuming its initial velocity to be zero.
KE=.5*2kg*v^2
18.6J=.5*2kg*v^2
v^2=18.6J/1kg
v^2=18.6m^2/s^2
v=sqrt 18.6m^2/s^2
v=4.31m/s
Good, except that the energy isn't 18.6 J. With the right energy, your calculation would give the correct result.
Using the definition of kinetic energy, determine the velocity of the cart after coasting the 95cm, assuming its initial velocity to be .38m/s
For this part of the question, I’m not sure how to apply the initial velocity change to .38m/s
Gravity does the same work as before, resulting in the same change in KE.
However the initial KE isn't zero. Calculate the initial KE, then add the change in KE to get the final KE. Then proceed as in your last calculation.
3. A mass of 45g is attached to a cart of mass 270g and suspended over a pulley of negligible mass and friction. The cart is placed on a ramp whose slope is just enough to compensate for the small frictional force acting on the cart. When the system is released, what will be the acceleration of the cart?
The force of the 270g cart I got to be 2.65N while the force of the 45g mass I got to be .441N. The net force on the system is 2.21N. So, using a=Fnet/m, I get
=2.21N/m1+m2
=2.21N/.315kg
=7m/s^2
Answer the same question if the cart is placed on a ramp making an angle of 4 degrees with horizontal, with the cart being pulled down the ramp, and if the frictional force is .024 times the normal force on the cart.
That would be correct if the cart wasn't being supported (e.g., if the two were hanging as masses on opposite sides of an Atwood system).
However the cart is being supported by the ramp, so its 2.65 N weight is countered by the normal force of the ramp and the friction.
Thus the net force on the system is just the .441 N.
If after adding an unknown mass to the cart on the 4 degree incline the acceleration of the system is 26.88cm/s^2, what is the unknown mass?
On these last 2 questions, again I’m a little confused on how to apply the 4 degree angle. And I see that once I answer the 2nd question above, I should just be able to apply a=Fnet/m to find the unknown mass.
4. What would be the orbital KE of a satellite of mass 830kg in circular orbit about a planet of mass 99*10^24kg, orbiting at a distance of 47000km from the center of the planet?
Satellite mass=830kg
Planet mass=99*10^24kg
Distance=47000km
Radius=51700km
G=6.67*10^-11Nm^2/kg^2
KE=1/2*m*(G*M)/r
=1/2*830kg*[(99*10^24kg*6.67*10^-11Nm^2/kg^2)]/47000km
=.415kg*(660kg*10^13/47000km)
That expression simplifies. If you do it correctly you should get the KE of the object in its orbit.
By how much would orbital KE change as the satellite moved from this orbit to a circular orbit of radius 51700km?
G=6.67*10^-11Nm^2/kg^2
I kinda got stuck when trying to solve this. I understand on the 2nd question that the KE will be negative whatever the difference in KE is between the 2 because as the orbit gets larger the KE decreases.
Just do the same thing you did before, but with the new radius. Then subtract your two results.
#$&*