Download Q3.2.a The gravitational force exerted by a planet on one of its

Q3.2.a
The gravitational force exerted by a planet on one of
its moons is 3e23 newtons when the moon is at a
particular location.
If the mass of the moon were three times as large,
what would the force on the moon be?
1) 1e23 N
2) 3e23 N
3) 6e23 N
4) 9e23 N
Q3.2.b
The gravitational force exerted by a planet on one of
its moons is 3e23 newtons when the moon is at a
particular location.
If the distance between the moon and the planet was
doubled, what would the force on the moon be?
1) 6e23 N
2) 3e23 N
3) 1.5e23 N
4) 0.75e23 N
5) 0.33e23 N
Q3.2.c:
1) < 0.949, 0.316, 0 >
2) < −0.949, −0.316, 0 >
3) < 0.447, 0.894, 0 >
We want to find the gravitational force on the 4) < 0.800, 0.600, 0 >
5) None of the above
planet by the star. What is the unit vector
r-hat?
Fixed star position: < 0.5e11, 1e11, 0 > m
Initial planet position: < 2e11, 1.5e11, 0 > m
Q3.2.c.alt.1:
Fixed star position:
< 0.5e11, 1e11, 0 > m
Initial planet position:
< 2e11, 1.5e11, 0 > m
Calculate the vector that points from
the star to the planet.
1) < 1e22, 1.5e22, 0 > m
2) < 1.5e11, 0.5e11, 0 > m
3) < -1.5e11, -0.5e11, 0 > m
4) < 2.5e11, 2.0e11, 0 > m
5) We don’t have enough
information to find the vector
Q3.2.c.alt.2
The relative position vector from the star to the planet 1) 0.50e11 m
2) 1.50e11 m
is:
3) 1.58e11 m
< 1.5e11, 0.5e11, 0 > m.
4) 2.00e11 m
What is the distance between the star and the planet? 5) 2.50e11 m
Q3.2.c.alt.3:
Relative position vector from star to
planet is < 1.5e11, 0.5e11, 0 > m
Distance from star to planet is 1.58e11 m
1) < 1, 0, 0 >
2) < 1, 1, 0 >
3) < 0.949, 0.316, 0 >
4) < 1.5e11, 0.5e11, 0 >
Find the unit vector pointing from the star 5) < 1.58e11, 0, 0 >
to the planet
Q3.2.d:
Distance from star to planet: 1.58e11 m
Star’s mass: 1e30 kg
Planet’s mass: 5e24 kg
G = 6.7e-11 N·m2/kg2
Calculate the magnitude of the
gravitational force that the
star exerts on the planet.
1) 1.34e-8 N
2) 2.68e-2 N
3) 1.34e22 N
4) 2.12e33 N
5) 5.3e55 N
Q3.2.e:
Calculate the gravitational force exerted
by the star on the planet
(remember that force is a vector)
1) < 1e22, 1.5e22, 0 > N
2) < 1.5e11, 0.5e11, 0 > N
3) < -1.5e11, -0.5e11, 0 > N
4) < 2.5e11, 2.0e11, 0 > N
5) < -1.27e22, -4.24e21, 0 > N
Q3.3.a
Mass of Mars: 6.4e23 kg; radius of Mars: 3.4e6 m
Mass of Earth: 6e24 kg; radius of Earth: 6.4e6 m
If the Mars rover measured the value of g on Mars, it would be:
1) 9.8 N/kg
2) less than 9.8 N/kg
3) more than 9.8 N/kg
Q3.4.a
The Earth has a mass of 6e24 kg. The Sun is much more massive; its mass is 2e30 kg.
Which of the following statements is correct?
1) The gravitational force on the Sun by the Earth is smaller in magnitude than the
gravitational force on the Earth by the Sun.
2) The gravitational force on the Sun by the Earth is exactly the same in magnitude as
the gravitational force on the Earth by the Sun.
3) Neither (1) nor (2) is correct.
Q3.4.b
You hold a tennis ball at rest
above your head, then open your
hand and release the ball, which
begins to fall.
At this moment, which statement
about the magnitudes of the
gravitational forces between the
Earth and ball is correct?
1) The force on the ball by the Earth is larger
than the force on the Earth by the ball.
2) The force on the ball by the Earth is
smaller than the force on the Earth by the
ball.
3) The force on the ball by the Earth is equal
to the force on the Earth by the ball.
4) There is not enough information to
determine this.
Q3.4.c
An alpha particle contains two
protons and two neutrons, and
has a net charge of +2e.
The alpha particle is 0.1 m away
from a single proton, which has
charge +e.
Which statement about the
magnitudes of the electric forces
between the particles is correct?
1) The force on the proton by the alpha particle
is equal to the force on the alpha particle by
the proton.
2) The force on the proton by the alpha particle
is larger than the force on the alpha particle by
the proton .
3) The force on the proton by the alpha particle
is smaller than the force on the alpha particle
by the proton.
4) There is not enough information to
determine this.
Q3.5.a:
The planet initially has a velocity of
1) < 5e28, -1e29, 0 > kg·m/s
< -1e4, 2e4, 0 > m/s.
2) < -5e28, 1e29, 0 > kg·m/s
What is the initial momentum of the planet?3) < 1e30, 5e24, 0 > kg·m/s
4) < -1e4, 2e4, 0 > kg·m/s
Star’s mass: 1e30 kg
Planet’s mass: 5e24 kg
5) < -2e-21, 4e-21, 0 > kg·m/s
G = 6.7e-11 N·m2/kg2.
Q3.5.b:
After 1 day (24*60*60 s), what is the 1) < 5.11e28, -9.97e28, 0 > kg·m/s
new momentum of the planet?
2) < -5.11e28, 9.97e28, 0 > kg·m/s
3) < -1.10e27, -3.66e26, 0 > kg·m/s
4) < -1e4, 2e4, 0 > kg·m/s
5) < -2e-21, 4e-21, 0 > kg·m/s
Q3.5.c: In step 1 we applied the Momentum Principle and updated position.
1) Relative position
After step 1 (result
vector
shown), which
2) Unit vector r-hat
quantities have
3) Force on planet
changed and must
by star
be recalculated for
4) All of these
step 2?
5) None of these
Q3.5d
We predicted the Earth’s
orbit around the Sun,
taking a step of 3 months.
Why was this prediction
so poor?
1) The Momentum Principle does not apply to
gravitational forces.
2) We neglected air resistance.
3) We did not use enough significant figures.
4) We assumed the force on the Earth was
constant in magnitude and direction over
three months.
3.6.a:
Which arrow best indicates the direction of
the net electric force on the blue negatively
charged object?
Q3.6.b:
Which arrow best indicates the direction of
the net electric force on the blue negatively
charged object?
Q3.11.a
A tennis ball falls for 1 second.
During this time the change in the
y component of the ball’s
momentum is
py = –0.6 kg m/s
What is the change in the y
component of the Earth’s
momentum?
1) –0.6 kg m/s
2) +0.6 kg m/s
3) zero because the ball does not exert
a force on the Earth
4) zero because the Earth’s momentum
can’t change
5) There is not enough information to
determine this.
Q3.11.b
When a ping pong ball collides
with a bowling ball, why is the
effect on the ping pong ball
more noticeable than the effect
on the bowling ball?
1) The momentum of the bowling ball does
not change.
2) The change in the bowling ball’s
momentum is less than the change in the
ping pong ball’s momentum.
3) The change in the bowling ball’s velocity
is less than the change in the ping pong
ball’s velocity.
Q3.11.c
A bowling ball is initially at rest, floating
in outer space. A ping pong ball moving
in the +z direction hits the bowling ball,
and bounces off it, traveling back in the
–z direction.
Consider a time interval t from slightly
before to slightly after the collision.
In this time interval, what is the sign
of pz for the system consisting of
both balls?
1) positive
2) negative
3) zero – no change in pz
Q3.11.d
A bullet of mass 0.04 kg traveling
horizontally at a speed of 800 m/s
embeds itself in a block of mass 0.5kg
that is sitting at rest on a very slippery
sheet of ice.
You want to find the speed of the block
just after the bullet embeds itself in the
block.
What should you choose as the
system?
1) the bullet
2) the block
3) the bullet and the block
Q3.11.e:
A bullet of mass 0.04 kg traveling horizontally at a speed of 800 m/s embeds
itself in a block of mass 0.50 kg that is sitting at rest on a very slippery sheet of ice.
Which equation will correctly give the final speed vf_BLOCK of the block?
1) (0.04 kg)*(800 m/s) = (0.50 kg) *vf_BLOCK
1) (0.04 kg)*(800 m/s) = (0.04 kg) *vf_BLOCK
1) (0.04 kg)*(800 m/s) = (0.50 kg) *vf_BLOCK + (0.04 kg)*(800 m/s)
1) (0.04 kg)*(800 m/s) = (0.54 kg) *vf_BLOCK
1) (0.04 kg)*(800 m/s) = (0.5 kg) *vf_BLOCK + (0.04 kg)*vf_bullet
Q3.11.f:
A space satellite of mass 500 kg has
velocity < 12, 0, –8 > m/s just before
being struck by a rock of mass 3 kg with
velocity < –3000, 0, 900 > m/s.
After the collision the rock’s velocity is
< 700, 0, –300 > m/s. Now what is the
velocity of the space satellite?
1) < –5100, 0, –400 > m/s
2) < –10.2, 0, –0.8 > m/s
3) < 10.2, 0, 0.8 > m/s
4) < –3688, 0, 1191 > m/s
5) < 3688, 0, –1192 > m/s