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: 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 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 is: < 1.5e11, 0.5e11, 0 > m. What is the distance between the star and the planet?

1) 0.50e11 m 2) 1.50e11 m 3) 1.58e11 m 4) 2.00e11 m 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 Find the unit vector pointing from the star to the planet

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 >

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 < -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?

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