Review Module – Physics
KINEMATICS IN ONE AND TWO DIMENSIONS
If acceleration is constant,
1
s = vo t + at2
2
If acceleration is varying,
ds
v=
dt
v2 = v2o + 2as
a=
dv
dt
v = vo + at
v dv = a ds
1. The position of the front bumper of a test car under microprocessor
control is given by x = 2.17 + 4.18t2 – 0.1t6, meters. Find its position
and acceleration at the instants when the car has zero velocity.
2. A car is traveling at the posted speed limit of 24.12 kph in a school
zone on dry pavement. The driver applies maximum braking force,
and the car decelerates at a constant rate of 7.92 m/s2 until coming
to a halt. Calculate the distance the car travels while decelerating.
ROTATIONAL MOTION
If angular acceleration is constant,
1
θ = ωo t + αt2
ω2 = ω2o + 2αθ
2
ω = ωo + αt
SITUATION. Initially, a ball has an angular velocity of 5.0 rad/s
counterclockwise. Considering a point in the circumference of the ball,
sometime later, it is at 5.5 radians relative to its initial position and the
ball now has an angular velocity of 1.5 rad/s clockwise.
13. What is the angular acceleration?
14. How much time did it take for the ball to have an angular velocity of
1.5 rad/s clockwise?
15. At some point, the angular velocity of the ball had to have been zero.
At what angle from its initial orientation did this occur and how long
did it take?
SITUATION. On Mars, a tennis ball is hit directly upward and returns to
the same level 8.5 seconds later. The acceleration due to gravity on Mars
is 0.379g and air resistance is negligible.
SITUATION. The angular position θ of a 0.36-m-diameter flywheel is
given by θ = 2t3, where t is in seconds.
3.
4.
5.
16. Find the distance that a particle on the flywheel rim moves from t1
= 2 s to t2 = 5 s.
17. Find the instantaneous angular velocity at t = 2 s.
18. Find the instantaneous angular acceleration at t = 2 s.
19. Find the total acceleration at t = 2 s.
How high above its original point did the ball go?
How fast was it moving just after being hit?
What is the total distance travelled by the ball?
SITUATION. A projectile is fired with an initial speed of 196 m/s at an
angle of 30° above the horizontal from the top of a cliff 98 m high.
6. Determine the time to reach maximum height.
7. Determine the maximum height above the base of the cliff reached
by the projectile.
8. Determine the total time it is in the air.
9. Determine the horizontal range of the projectile.
CURVILINEAR MOTION
Tangential Acceleration, at
at = same as rectilinear motion
Normal Acceleration, an
v2
an =
r
SITUATION. A car travels over the crest of a hill at 10 m/s. The radius at
the crest is 12 m.
ERRATIC MOTION
SITUATION. An object which started from rest is travelling from point A
to point C for 15 seconds. During the first 10 seconds, the object has a
constant acceleration of 4 m/s2 and on the last 5 seconds, the object
decelerated at a constant rate of -2 m/s2.
20. Determine the velocity of the object after 15 seconds.
21. Determine the total distance travelled after 15 seconds.
KINETICS: FORCE AND ACCELERATION
22. A 50-kg crate rests on a horizontal surface for which the coefficient
of kinetic friction is μk = 0.30. The crate is subjected to a 400-N
towing force directed at 30° above the horizontal. Determine the
acceleration of the crate.
SITUATION. The 1200-kg Trailer B is hitched to the four-wheel-drive
truck A of mass 1800 kg. The traction force developed at the wheels is
16,000 N. Neglect rolling resistance of the trailer.
10. Determine the force exerted by the car seat on a 60-kg passenger.
11. Determine the minimum speed required for the passenger to feel
momentarily “weightless.”
12. At some point on the hill, the car traveling at 10 m/s accelerates
tangentially to a constant rate of 6 m/s2. If the radius of curvature is
50 m, determine the total acceleration experienced by the car at that
instant.
23. Determine the maximum possible acceleration.
24. Determine the tensile force in the trailer hitch.
KINETICS: IMPULSE AND MOMENTUM
SITUATION. A 500-g tetherball is
moving along a horizontal circular path at
a constant speed of 4 m/s.
Impulse-Momentum Theorem
The impulse of the net external force on a particle during a time interval
equals the change in momentum of that particle during that interval:
25. Find the angle θ that the 1.8-m
cord forms with the pole.
26. Find the tension in the cord.
KINETICS: WORK AND ENERGY
Work-Energy Theorem
Work done by the net force on a particle equals the change in the
particle’s kinetic energy.
1
∑ Work = m(v22 - v21 )
2
Work Due to External Force
Work Due to Weight
W = Fd
W = mgh
Work Due to Spring
W = 2 kx2
1
Power – the rate at which work is done (in Watts or Joules/sec)
Work
P = time = Fv
SITUATION. A 1.50 kg book is sliding along a rough horizontal surface.
At point A, it is moving at 3.20 m/s, and at point B it has slowed to
1.25 m/s.
27. How much total work was done on the book between A and B?
28. If -0.85 J of total work is done on the book from B to C, how fast is
it moving at point C? What if the total work is +0.85 J?
29. How much work needs to be done on the book from B to C to stop
it from moving?
SITUATION. A spring, with spring constant k = 30,000 N/m, is used to
stop a 50-kg package which is moving down a 20° incline. When the
package is 8 m from the spring, its velocity is 2 m/s. Coefficient of kinetic
friction is 0.20.
∑ Fnet (∆t) = m(∆v) – for constant force only
t
∫t 2 F dt = ∆(mv) – for varying force
1
Law of Conservation of Momentum
The total linear momentum for a system of particles remains constant
during the time period t1 to t2.
∑ mi (vi )1 = ∑ mi (vi )2
SITUATION. A 50-kg body initially at rest is acted upon by a constant
force of 80 N inclined at an angle θ above the horizontal for 5 seconds,
after which an opposite horizontal force of 45 N is applied.
33. If θ = 0°, what is the velocity after 10 seconds from rest?
34. If θ = 0°, what total time in seconds should the 45-N force be
applied so that the body comes to rest?
35. If θ = 20°, what is the velocity after 5 seconds from rest?
SITUATION. The 15,000-kg boxcar A is coasting at 1.5 m/s on the
horizontal track when it encounters a 12,000-kg tank car B coasting at
0.75 m/s toward it. The cars collide and couple together.
36. Find the speed of both cars just after the coupling.
37. Find the average force between them if the coupling takes place in
0.8 s.
KINETICS: IMPACT AND COLLISION
The ratio of the restitution impulse to the deformation impulse is called
the coefficient of restitution, e.
(vB )2 - (vA )2 (vA )2 - (vB )2
e==
(vB )1 - (vA )1 (vB )1 - (vA )1
When e = 1, energy loss is equal to 0 and the impact is called perfectly
elastic impact.
When e = 0, energy loss is maximum, and the impact is called perfectly
plastic/inelastic impact.
SITUATION. A 20-kg object A travelling at 4 m/s collides head on with a
10-kg object B travelling at 2 m/s opposite its direction. Determine the
velocity of each object just after the impact given the following:
30. Just before striking the spring, determine the work due to weight.
31. Just before striking the spring, determine the work due to external
force.
32. Determine the maximum deformation of the spring.
38. The collision is perfectly elastic.
39. The collision is perfectly inelastic.
40. The coefficient of restitution is 0.45.
“It’s going to be hard, but
hard does not mean impossible.”