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?
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) 1e23 N
2) 3e23 N
3) 6e23 N
4) 9e23 N
1) 6e23 N
2) 3e23 N
3) 1.5e23 N
4) 0.75e23 N
5) 0.33e23 N
Q3.2.c:
Fixed star position: < 0.5e11, 1e11, 0 > m
Initial planet position: < 2e11, 1.5e11, 0 > m
We want to find the gravitational force on the planet by the star. What is
the unit vector r-hat?
1) < 0.949, 0.316, 0 >
2) < −0.949, −0.316, 0 >
3) < 0.447, 0.894, 0 >
4) < 0.800, 0.600, 0 >
5) None of the above
Q3.2.c.alt.1:
Fixed star position:
< 0.5e11, 1e11, 0 > m
Initial planet position:
< 2e11, 1.5e11, 0 > m
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
Calculate the vector that points from the star to the planet.
Q3.2.c.alt.2
The relative position vector from the star to the planet is:
< 1.5e11, 0.5e11, 0 > m.
What is the distance between the star and the planet?
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
Find the unit vector pointing from the star to the planet
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1) 0.50e11 m
2) 1.50e11 m
3) 1.58e11 m
4) 2.00e11 m
5) 2.50e11 m
1) < 1, 0, 0 >
2) < 1, 1, 0 >
3) < 0.949, 0.316, 0 >
4) < 1.5e11, 0.5e11, 0 >
5) < 1.58e11, 0, 0 >
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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
1) 1.34e-8 N
2) 2.68e-2 N
3) 1.34e22 N
4) 2.12e33 N
5) 5.3e55 N
Calculate the magnitude of the
gravitational force that the
star exerts on the planet.
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?
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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.
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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
< -1e4, 2e4, 0 > m/s.
What is the initial momentum of the planet?
Star’s mass: 1e30 kg
Planet’s mass: 5e24 kg
G = 6.7e-11 N·m2/kg2.
1) < 5e28, -1e29, 0 > kg·m/s
2) < -5e28, 1e29, 0 > kg·m/s
3) < 1e30, 5e24, 0 > kg·m/s
4) < -1e4, 2e4, 0 > kg·m/s
5) < -2e-21, 4e-21, 0 > kg·m/s
Q3.5.b:
After 1 day (24*60*60 s), what is the new momentum of the
planet?
1) < 5.11e28, -9.97e28, 0 > kg·m/s
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.
After step 1 (result shown),
which quantities have changed
and must be recalculated for
step 2?
1) Relative position vector
2) Unit vector r-hat
3) Force on planet
by star
4) All of these
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? 7
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Q3.6.b:
Which arrow best indicates the direction of the net electric force on the blue
negatively charged object? 7
Q3.6c
A sodium ion exerts a force of 3e–8 N on a proton that is a
distance d away. If the sodium ion were moved to a distance 3d
from the proton, what would the magnitude of the force on the
proton be?
1) 9e–8 N
2) 3e–8 N
3) 1e–8 N
4) 0.33e–8 N
5) 0.11e–8 N
Q3.11.a
A tennis ball falls for 1 second. During this time the change in 1) –0.6 kg m/s
the y component of the ball’s momentum is
2) +0.6 kg m/s
3) zero because the ball does not exert a force on the Earth
py = –0.6 kg m/s
4) zero because the Earth’s momentum can’t change
5) There is not enough information to determine this.
What is the change in the y component of the Earth’s
momentum?
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.
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.
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
What should you choose as the
system?
1) the bullet
2) the block
3) the bullet and the block
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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
pi = pi
500 * 12, 0, -8 + 3* -3000, 0, 900 = 3* 700, 0, -300 + 500v f
-3000, 0, -1300 - 2100, 0, -900
= vf
500
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