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Transcript
Chapter 2 Review Questions
Knowledge
1.
Name
Symbol
Most Common Unit
momentum
kg•m/s
change in momentum
kg•m/s
impulse
impulse
Formula
N•s
Newton’s third law
N
conservation of
momentum
kg•m/s
work
W
J
kinetic energy
Ek
J
gravitational potential
energy
Ep(grav)
J
W = Fd
Ep = mgh
2. Mass and velocity are the two quantities that must be known to determine momentum.
3. Momentum is the quantity of motion.
4. Impulse is the product of the amount of force exerted on an object and the time interval
during which the force is exerted.
5. The seat belt increases the amount of time for a change in momentum; therefore, less force
is exerted to change the momentum.
6. The force exerted by object 1 on object 2 is equal but opposite in direction to the force
exerted by object 2 on object 1. For example, when you push on a wall with a force of 10 N,
the wall exerts an equal force of 10 N on you.
7. Due to Newton’s third law, the ball exerts an equal but opposite force on you that propels
you into the water.
8. If the net force acting on a system is zero, the total momentum before an interaction is equal
to the total momentum after the interaction.
9. The three types of interactions are hit and stick, hit and rebound, and explosion.
Applying Concepts
10. Units for Impulse
N•s = (kg•m/s2)•s
= kg•m/s
Units for Change in Momentum
kg•m/s
Therefore, the units for impulse is equal to the units for change in momentum.
11. Let east be the positive direction.
a.
The initial momentum is 1.7 × 102 kg•m/s [E].
b.
The final momentum is 0.
c.
The change in momentum is 1.7 × 102 kg•m/s [W].
d.
The impulse provided by the foam barrier is 1.7 × 102 N•s [W].
e.
The force provided by the foam barrier is 1.4 × 103 N [W].
f.
g.
The acceleration of the combined mass is 22 m/s2 [W].
m = 66 kg
v = 2.6 m/s
Ek=?
The initial kinetic energy is 2.2 × 102 J.
h.
i.
12. a.
Because the final velocity is 0, the final kinetic energy is 0.
Ek(initial) = 223.08 J
Ek(final) = 0
Ek = ?
The change in kinetic energy is - 2.2 × 102 J.
F = 1.5 × 103 N
W =
d = 0.15 m
=
W=?
=
=
Fd
(1.5 × 103 N)(0.15 m)
225 J
2.3 × 102 J
The work done by the foam barrier is 2.3 × 102 J.
b. The magnitude of the work done is equal to the change in kinetic energy.
13. Let north be the positive direction.
a.
The initial momentum of the car is 1.5 × 104 kg•m/s [N].
b.
The final velocity of the truck is 13 m/s [N].
c.
The change in momentum of the car is 1.5 × 104 kg•m/s [S].
d.
e.
f.
The force exerted by the truck in stopping the car was 1.0 × 105 N [S].
F = 1.0 × 105 N
W = Fd
d = 0.060 m
= (1.0 × 105 N)(0.060 m)
W=?
= 6.0 × 103 N•m
= 6.0 × 103 J
The work done on the car by the truck is 6.0 × 103 J.
The car initially had kinetic energy. After the collision, all of its kinetic energy converted
into kinetic energy of the other vehicle as well as heat energy, sound energy, and energy
to transform the shape of the car.
14. m1 = 900 kg
m2 = 1500 kg
The final velocity of the combined mass is 8.0 m/s [N].
15. Let the original (forward) direction of the ball be the positive direction.
a.
The change in momentum of the ball is 5.50 kg•m/s [backward].
b.
The force provided by the wall on the ball is 25 N [backward].
16. No, a primary collision will not cause injury to a passenger in a vehicle. The primary collision
is between the vehicle and the obstacle. The secondary and tertiary collisions will cause
injury to the occupant.
17. Let north be the positive direction.
a. m1 = 4.0 kg
m2 = 2.0 kg
The total momentum before is equal to the total momentum after; therefore, momentum is
conserved.
b.
The total kinetic energy before does not equal the total kinetic energy after. This does not
mean that energy is not conserved. It means that some of the kinetic energy before the
collision converted into some other form of energy. Specifically, some of the kinetic energy
was used to change the shape of the balls so they could stick together and some of the kinetic
energy was converted into heat and sound energy.
18. a.
Since
and since the
is the same in both cases, F must decrease as
increases.
Letting your hand “give” increases the time, thus making catching the ball less painful
because of the decreased force.
b. Let the original (forward) direction of the ball be the positive direction.
Letting Arms “Give”
The force exerted by his arm on the ball over 1.2 s is 7.0 N [backward].
Rigid Arms
The force exerted by his arm on the ball over 0.40 s is 21 N [backward].
19. Let north be the positive direction.
a. m1 = 1500 kg
The initial momentum of the car is + 3.8 × 104 kg•m/s [N].
The initial momentum of the minivan is 1.3 × 104 kg•m/s [S].
b.
The total momentum of the two vehicles after the collision is 2.4 × 10 4 kg•m/s [N].
c.
The final velocity of the two vehicles after the collision is 9.0 m/s [N].
20. a.
b.
The area is equal to impulse, since the area under the graph is equal to
c. Since the question is asking for final speed (a scalar), the vector notation is dropped.
The final speed of the mass is 3.33 m/s.
Science 20 © 2006, Alberta Education