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GPHS Physics
Final Exam Review KEY
This year’s final will be short. 10 MC questions and 2 or three FR questions. It will cover content from
the entire year, and it will try to de-emphasize math. It will be worth about the same as our other tests.
Chapter 1: 1 D Motion
1. Robin Hood shoots and arrow straight up in the air. It leaves his bow with a speed of 50 m/s.
(Hint: You should be able to do all of these with only a few simple mental calculations)
a. How fast will it be going when it has reached its highest point?
The speed will be 0 m/s when it stops at the top.
b. How long will it be ascending for?
The arrow loses 10 m/s each second, so it will take 5 seconds till it is stopped
momentarily at the top.
c. From the time it is shot, how long will it take the arrow to land?
10 seconds. 5 going up and 5 going down.
d. How fast will it be moving when it lands?
50 m/s. The speed of a projectile is symmetrical. It is going the same speed at a given
altitude whether it is going up or down.
e. What is the arrow’s acceleration at the apex of its flight?
a=g for the entire flight. Its speed is zero at the top but is acceleration is constant for
the whole flight.
f. When, if ever, is its acceleration greatest?
See above.
Chapter 2: 2 D Motion and projectiles
2. A soccer ball is kicked from the top of a 20 m tall cliff. It has an initial velocity of 25 m/s and
leaves the ground at a 37° incline to the horizontal.
a. Write an equation for the horizontal distance traveled at any time, t, after it is kicked.
Using x(t) = x0 + v0t + ½at2, you get x(t)= 20 m/s t
b. Write an equation for the height of the ball above a point at the bottom of the cliff at
any time, t, after it is kicked.
Using the same equation you get y = 20 m +15 m/s t - 5 m/s/s t2
c. How fast is the ball moving at its highest point?
The speed will be 20 m/s, since vx is constantly 20 m/s and vy is 0 m/s at the apex.
d. How far will the ball travel before it lands
First solve the equation from (b) by factoring to find when the ball lands.
0 = 20 + 15 t – 5t2 = 4 + 3t – t2 = t2 – 3t – 4 = (t – 4)(t + 3)  t = 4 seconds
Then use your equation from a to find x(4) = 80 m
e. How fast will it be going when it lands?
𝑣 = √𝑣𝑥 2 + 𝑣𝑦 2 = √202 + 252 =32 m/s
Chapter 3: Newton’s Laws
3. A 150 kg caveman statue rests on a rough horizontal surface where the coefficients of friction
are µk=.40 and µs=.60 respectively. Pushing with all his might, student applies a horizontal force
of 900 N to the statue.
Fn
a. Draw an FBD for the statue.
f
F
W=mg
b. Determine the minimum force required to make the statue move, assuming it began at
rest.
Fmin = Fsmax = μsmg = .6(150kg)10m/s/s = 900N
c. Determine the acceleration of the statue if 900 N force described above is applied.
The acceleration would be zero because the net force is zero. But if you assume an
additional nudge to get the statue moving then
a = F / m = (900 N – fk) / 150 kg = (900 N – μkmg) / 150 kg = 300 N / 150 kg = 2 m/s/s
4.
d. Redraw the FBD for the statue if the force were applied at downward angle of 25°.
Fn (increases)
f (decreases)
F (angle changes, magnitude doesn’t)
W=mg (no change)
a. Give 2 reasons why the acceleration of the statue would decrease if the force were
applied downward.
1) The horizontal component of the applied force decreases
2) The normal force increase which increases friction.
Chapter 4: Uniform Circular motion
5. A car of mass 1500 Kg drives around a sharp corner of radius 25 m. The coefficient of static
friction between the tires and the road is .80.
a. In what direction is the car accelerating as it turns the corner?
Acceleration is always centripetal, or towards the center of the circle for any object in
UCM. Ever. Anywhere. Period. Exclamation point.
b. What is the magnitude of that acceleration?
ac = v2/r . You will have to find vmax in part (c). ac = 8 m/s/s
c. What is the fastest the car can go without slipping?
Fc = mac  μmg = mv2/r  v2 = μrg  v = 14.1 m/s
d. How would this maximum speed change for a 2500 kg truck, assuming an identical
coefficient of friction?
Mass Kirchunks, as you can see above.
Chapter 5 Energy
6. A spring is compressed 5 cm when a 50 N force is applied. The same spring is them compressed
20 cm and used to launch a 200 gram golf ball vertically.
a. Describe the transformations of energy in this situation
Elastic Potential  Kinetic  Gravitational potential
b. Find k, the spring constant for the spring
F = -kx  50N = -k (-.05 m)  k = 1000 N/m
c. How fast will the ball be going at launch?
Ue=K  ½ kx2 = ½ mv2  v2 = kx2/m  v = 6.25 m/s
d. How high will the ball go?
mgh = ½ kx2  h = kx2 / (mg) = .625 m
Chapter 6: Momentum and Impulse
7. A 1.25 kg physics cart sits at rest on a frictionless track. Another physics cart, this one with a
mass of 750 grams is moving 5 m/s along the track when it collides with the first cart.
Afterwards, the two carts move apart. The heavier cart is found to be moving 4 m/s after the
collision.
a. Find the impulse experienced by the 1.25 kg cart
J = mΔv = 1.25 kg (4 m/s) = 5 kg m/s
b. Find the final speed of the 750 g cart.
pi=pf  .750 kg (5 m/s) = 1.25 kg (4 m/s) + .750 vf  3.75 kg m/s = 5 kg m/s + .750 vf
vf = -1.25 kg m/s / .75 kg  vf = 1.67 m/s BACKWARDS
c. How would this collision have looked different if the two carts collided inelastically?
The two carts would be stuck together afterwards, the impulse experienced by each
would be less, and more kinetic energy would be lost to heat and destruction.
Chapter 7: Gravity and Orbits
8. Use Newton’s Laws of gravity and motion to show that the acceleration due to gravity for any
object at earth’s surface must be 9.81 m/s/s as we have observed.
F = ma  GMm / r2 = ma  a = GM/r2 = (6.11 x 10-11)(5.97 x 1024 kg) / (6.67 x 106 m)2
a = 9.81 m/s/s
Chapter 8: Electricity
9. A positively charged plastic rod is brought near a neutrally charged metal ball hanging from a
thin string. The ball is mysteriously attracted
to the rod.
a. Sketch the movement of charges that
causes this attraction
b. Why would this not work if the rod
were metal?
A metal rod would allow the charges built up on the rod to disperse through your hand
and into the ground. It is un natural to keep the excess charge together.
c. Why would this not work if the ball were plastic?
If the ball were plastic, charges would not be free to move about inside the ball since
plastic is an insulator.
10. Two small point charges of charges +3µC and -6µC are separated by a distance of 5 cm. There
are no other charges around.
a. Sketch the two charges, and the electric field surrounding
them.
b. Find the force between the two charges.
F = 9 x 109 N m2/C2 (3 x 10-6 C) (-6 x 10-6 C) / .05 m2
F = 64.8 N the force is attractive.
c. What are the magnitude and direction of the electric field at point P which lies half way
between the two charges?
First, consider the direction. The E field tells you the force that a positive charge would
feel if it were at a point in space. Half way in between, the positive charge would push
our test charge right, and the negative would pull a test charge right, so these two parts
will add together.
E = [9 x 109 N m2/C2 (3 x 10-6 C) / (.05 m)2] + [9 x 109 N m2/C2 (-6 x 10-6 C) / (.05 m)2]
E = 10,800,000 N/C + 21,600,000 N/C = 32.4 MN/C
d. What is the electric potential at P?
Since potential is a scalar, we do not need to stop and think about the direction. We can
just let the signs of the charges do the work for us.
V = [9 x 109 N m2/C2 (3 x 10-6 C) / (.05 m)] + [9 x 109 N m2/C2 (-6 x 10-6 C) / (.05 m)]
V = 540,000 V – 10,800,000 V = -540,000 V