Unit 4 Test Flashcards by dhruv patel

(A) It is elastic because momentum is conserved.
(B) It is elastic because kinetic energy is conserved.
(C) It is inelastic because momentum is not conserved.
(D) It is inelastic because kinetic energy is not conserved.
(E) More information is needed to determine whether the collision is elastic or inelastic

(B) It is elastic because kinetic energy is conserved.

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(A) vf < v0, because some of the cart-bag system’s mechanical energy is dissipated.
(B) vf < v0, because the kinetic energy of the bag decreases.
(C) vf > v0, because the kinetic energy of the bag decreases.
(D) vf > v0, because the linear momentum of the bag is added to the linear momentum of the cart.
(E) vf = v0, because the bag is moving perpendicular to the cart.

(A) vf < v0, because some of the cart-bag system’s mechanical energy is dissipated.

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Students perform an experiment with multiple trials using blocks that can slide with negligible friction on a straight,
horizontal track. In each trial, a block of mass m1 slides with speed v1 on the track and collides with a stationary
block of mass m2 . The blocks stick together and move with speed vf after the collision. In each trial, m1 and v1
are kept constant but the mass m2 of the stationary block is varied, and vf is recorded. The students graph vf as a
function of m2. Which of the following could represent a best-fit curve to the data collected by the students?

(A) going down concave down
(B) going down linear
(C) going down concave up
(D) going up linear

(A) going down concave down

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Two objects of equal mass undergo a series of collisions. For each collision, a graph is created of the momenta of
both objects as functions of time. Which of the following graphs represents an elastic collision?

(A)
(B)
(C)
(D)

(B)

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(A) 54J
(B) 90 J
(C) 108 J
(D) 162 J

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(A) Yes, because the fact that the velocity of Object 1 changed direction during the collision means that the collision was not elastic
(B) Yes, because conservation of momentum can be used to determine the final speed of Object 2, and the total kinetic energy of the two-object system after the collision can be calculated
(C) No, because the final speed of Object 2 cannot be determined without knowing whether the collision was elastic or inelastic
(D) No, because the collision can be elastic for many different values of the final speed of Object 2

(B) Yes, because conservation of momentum can be used to determine the final speed of Object 2, and the total kinetic energy of the two-object system after the collision can be calculated

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(A) Cart A is moving to the left and Cart B is moving to the right
(B) Cart A is stationary and cart B is moving to the right
(C) Both carts are moving to the right at the same speed
(D) Both carts are moving to the right, but they travel at different speeds

(D) Both carts are moving to the right, but they travel at different speeds

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(A) 1/3
(B) 1/2
(C) 2/3
(D) 1

(D) 1

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(A) Immediately after Collision 1. During Collision 1, Cart Y gains an among of kinetic energy equal to Ki, while in Collision 2, Cart Y gains an amount of kinetic energy less than Ki
(B) Immediately after Collision 1. During Collision 1, Cart Y gains an amount of kinetic energy greater than Ki, while in Collision2, Cart Y gains an amount of kinetic energy equal to Ki.
(C) Immediately after Collision2. In Collision 1, the two-cart system loses more kinetic energy than in Collision 2.
(D) Immediately after Collision2. When the carts stick together, Cart X transfers more energy to Cart Y than when the carts bounce off of one another

(A) Immediately after Collision 1. During Collision 1, Cart Y gains an among of kinetic energy equal to Ki, while in Collision 2, Cart Y gains an amount of kinetic energy less than Ki

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(A) 2/3 J
(B) 1 J
(C) 2 J
(D) 6 J

(A) 2/3 J

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(A) At position Q only
(B) At position R only
(C) At position S only
(D) At position Q, R, and S

(A) At position Q only

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(A) 1.5 m/s
(B) 4.0 m/s
(C) 4.5 m/s
(D) 5.7 m/s

(D) 5.7 m/s

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(A)
If the collision between block 1 and the wall is elastic, and the two blocks have an elastic collision, then
the sum of the kinetic energy of blocks 1 and 2 before block 1 makes contact with the wall will be equal
to the sum of the kinetic energy of blocks 1 and 2 after the collision between the blocks.
(B)
If the collision between block 1 and the wall is elastic, and the two blocks have an inelastic collision,
then the sum of the kinetic energy of blocks 1 and 2 before block 1 makes contact with the wall will be
equal to the sum of the kinetic energy of blocks 1 and 2 after the collision between the blocks.
(C)
If the collision between block 1 and the wall is not elastic, and the two blocks have an inelastic collision,
then the sum of the kinetic energy of blocks 1 and 2 before block 1 makes contact with the wall will be
less than the sum of the kinetic energy of blocks 1 and 2 after the collision between the blocks
(D)
If the collision between block 1 and the wall is not elastic, and the two blocks have an inelastic collision,
then the momentum of block 1 before it makes contact with the wall will be equal to the sum of the
momenta of the two blocks after the collision between the blocks.
(E)
If the collision between block 1 and the wall is not elastic, and the two blocks have an inelastic collision,
then the momentum of block 1 before it makes contact with the wall will be equal to the sum of the
momenta of the two blocks after the collision between the blocks.

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(A) Momentum is always conserved, and kinetic energy may be conserved.
(B) Kinetic energy is always conserved, and momentum may be conserved.
(C) Momentum is always conserved, and kinetic energy is never conserved.
(D) Both momentum and kinetic energy are always conserved.
(E) Neither momentum nor kinetic energy is conserved.

(A) Momentum is always conserved, and kinetic energy may be conserved.

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(A) Zero
(B) 0.5 m/s to the left
(C) 1 m/s to the right
(D) 2.4 m/s to the left
(E) 2.5 m/s to the right

(A) Zero

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(A) 1.0 kg * m/s
(B) 3.5 kg * m/s
(C) 5.0 kg * m/s
(D) 7.0 kg * m/s
(E) 5.5sqrt(5) kg * m/s

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(A) The total mechanical energy is constant.
(B) The total potential energy is constant.
(C) The total kinetic energy is constant.
(D) The total linear momentum is constant.
(E) It is in static equilibrium.

(D) The total linear momentum is constant.

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(A) 0
(B) v/6
(C) v/3
(D) v/2
(E) 6v

(B) v/6

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(A) Mv/m
(B) (M + m)v / m
(C) (M-m)v /m
(D) mv / (M)
(E) mv (M-m)

(B) (M + m)v / m

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(A) v/3
(B) v/2
(C) 2v/3
(D) 3v/2
(E) 2v

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(A) 4/3 m/s
(B) 8/3 m/s
(C) 4m/s
(D) 8m/s
(E) 24m/s

(A) 4/3 m/s

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A 2 kg ball collides with the floor at an angle ⯑ and rebounds at the same angle and speed as shown above. Which
of the following vectors represents the impulse exerted on the ball by the floor?

(A) Arrow pointing down
(B) Arrow pointing down right
(C) Arrow pointing right
(D) Arrow pointing up right
(E) Arrow pointing up

(E) Arrow pointing up

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(A) The slope would stay the same.
(B) The slope would be one-half as great.
(C) The slope would be one-quarter as great.
(D) The slope would be twice as great.
(E) The slope would be four times as great.

(A) The slope would stay the same.

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(A) The magnitude of change in momentum is the same for each puck, but Puck A has more momentum initially
(B) The magnitude of change in momentum is the same for each puck, but Puck B has more momentum initially
(C) The magnitude of change of momentum is the same for each puck because each puck has the same momentum initially
(D) The magnitude of change in momentum of Puck A is greater than that of Puck B because Puck A has a greater initial momentum
(E) The magnitude of change of momentum of Puck A is less than that of Puck B because Puck A has greater mass

(A) The magnitude of change in momentum is the same for each puck, but Puck A has more momentum initially

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(A) v0/2 (B) v0/3 (C) 7v0/5 (D) 3v0/2 (E) 2v0

(E) 2v0

Two balls are on a frictionless horizontal tabletop. Ball X initially moves at 10 meters per second, as shown in Figure I above. It then collides elastically with identical ball Y, which is initially at rest. After the collision, ball X moves at 6 meters per second along a path at 53° to its original direction, as shown in Figure II above. Which of the following diagrams best represents the motion of ball Y after the collision? (A) v = 0 (B) v = 10m/s (C) v = 4m/s theta = 37 (D) v = 8m/s theta 37 (E) v = 8m/s theta = 53

(D) v = 8m/s theta 37

(A) v/3 (D) v/2 (C) 2v/3 (D) v (E) 2v

(A) The mass of block 1 is accurate, but the mass of block 2 is actually less massive than recorded (B) both mass values are accurate, but v0 is actually larger than recorded (C) The motion sensor was positions too close to block 1 (D) The final velocity recorded was taken too long after the collision (E) The data recorded is accurate within acceptable experimental error

(D) The final velocity recorded was taken too long after the collision

(A) F2 = 2F1 (B) F2 = 4F1 (C) F2 = 8F1 (D) F2 = 16 F1

(A) F2 = 2F1

(A) The magnitude of the momentum is increasing (B) The magnitude of the momentum remains the same (C) The magnitude of the momentum is decreasing. (D) The answer cannot be determined without knowing the cart's mass and the mass of water that leaks through the hole each second.

(A) 0 (B) sqrt(2gD) (C) sqrt(4gD) (D) sqrt(8gD)

(D) sqrt(8gD)

A semicircular path connects three points labeled X, Y, Z, as shown in the figure. A particle is initially at point X and moves along the path with constant speed towards Z. When particle is at point Y, as shown in the figure, which of the following diagrams correctly represents the directions of the velocity v and the momentum p of the object? (A) (B) (C) (D)

(A) velocity must be tangent to the circular path and the momentum p is parallel to the velocity

(A) The frame-sphere system (B) The frame-Earth system (C) The sphere-Earth system (D) None of these two-object systems have a constant momentum.

(D) None of these two-object systems have a constant momentum.

(A) It is greater than the speed of Object 1 (B) It is equal to the speed of Object 1 (C) It is less than the speed of Object 1 (D) It could be greater than, equal to, or less than the speed of Object 1 depending on the directions of the momenta

(A) It is greater than the speed of Object 1

(A) 1.2 kg * m/s (B) 12kg * m/s (C) 24kg * m/s (D) 240 kg * m/s (E) The change in momentum of the block cannot be determined without knowing the block's mass

(A) 1.2 kg * m/s

(A) 5 N (B) 10 N (C) 20 N (D) 25 N (E) 50 N

(E) 50 N

(A) 1/4 p0 (B) 1/2 p0 (C) p0 (D) 2p0

(A) 1.0 kg * m/s (B) 2.5 kg * m/s (C) 4.5 kg * m/s (D) 7.5 kg * m/s

(D) 7.5 kg * m/s

(A) 16 kg * m/s (B) 24 kg * m/s (C) 48 kg * m/s (D) 64 kg * m/s

(A) 1 (B) 2 (C) 4 (D) 16

(B) 2

(A) 2p0 (B) 2p0 / sqrt(2) (C) p0 (D) p0 / sqrt(2)

(B) 2p0 / sqrt(2)

(A) 0.5 kg * m/s (B) 2.5 kg * m/s (C) 3.5 kg * m/s (D) 5.0 kg * m/s

(B) 2.5 kg * m/s

(A) J1 = -J2 (B) J1 = J2 (C) J1 < J2 (D) J1 > J2

(A) J1 = -J2

(A) k(t2^2 - t1^2) (B) kt2^2 - mgsin(theta)(t2) (C) k(t2^2 - t1^2) - mgsin(theta)(t2-t1) (D) k(t2^2 - t1^2) -mg(t2 - t1)

(A) The launcher will remain in place because there is no horizontal force acting on the launcher. (B) The launcher will move at a constant speed because the water is ejected at a constant speed. (C) The launcher will move at a constant acceleration because the system of the launcher and remaining water experiences the same impulse during any time interval of length change in t. (D) The launcher will move at an increasing acceleration because the mass of the system of the launcher and remaining water is decreasing, and the system experiences the same impulse during any time interval of length change in t.

(D) The launcher will move at an increasing acceleration because the mass of the system of the launcher and remaining water is decreasing, and the system experiences the same impulse during any time interval of length change in t.

(A) Use balls of the same mass and with the same initial speed but with different amounts of elasticity and repeat this procedure for each ball. If balls that rebound with less speed cause greater movement of the block, this is consistent with the prediction of the theorem. (B) Use balls of different masses but with the same amount of elasticity and the same initial speed and repeat this procedure for each ball. If smaller-mass balls cause more movement of the block, this is consistent with the prediction of the theorem. (C) Use identical balls with the same initial speed but blocks of different masses and repeat this procedure for each block. If the ball rebounds with more speed when it pushes a large block than when it pushes a smaller block, this is consistent with the prediction of the theorem. (D) Use balls with different masses but with the same amount of elasticity and same initial speed and repeat this procedure for each ball. If the balls that rebound with more speed cause less movement of the block, this is consistent with the prediction of the theorem. (E) Use balls of the same mass and with the same amount of elasticity but roll them toward the block at different speeds and repeat this procedure for each speed. If the balls traveling at higher speed cause less movement of the block, this is consistent with the prediction of the theorem

(C) Use identical balls with the same initial speed but blocks of different masses and repeat this procedure for each block. If the ball rebounds with more speed when it pushes a large block than when it pushes a smaller block, this is consistent with the prediction of the theorem.

(A) Yes, because the change in momentum is equal to the area under the curve of the force-versus-time graph. (B) Yes, because the change in momentum is equal to the slope of the force-versus-time graph. (C) No, because the initial momentum is unknown and therefore the change in momentum cannot be determined. (D) No, because the force is not constant and therefore the change in momentum cannot be determined

(A) Yes, because the change in momentum is equal to the area under the curve of the force-versus-time graph.

(A) F0w[cos(wt) i + sin(wt) j ] (B) F0w[cos(wt) i - sin(wt) j ] (C) F0/w [-cos(wt) i + sin(wt) j ] (D) F0/w [(1 -cos(wt) )i + sin(wt) j ]

(D) F0/w [(1 -cos(wt) )i + sin(wt) j ]

(A) W1 < W2 and J1 < J2 (B) W1 < W2 and J1 > J2 (C) W1 > W2 and J1 < J2 (D) W1 > W2 and J1 > J2

(A) W1 < W2 and J1 < J2

(A) zero (B) bF0 (C) F0/b (D) -F0/b

(A) 3kt (B) 6kt (C) 12kt (D) 2kt^3 (E) 3kt^3

(D) 2kt^3

(A) Increases, Decreases, Increases (B) Decreases, Increases, Remains the same (C) Increases, Decreases, Remains the same (D) Decreases, Remains the same, Decreases

(B) Decreases, Increases, Remains the same

(A) 6kg * m/s (B) 12kg * m/s (C) 17kg * m/s (D) 30kg * m/s

(B) 12kg * m/s

(A) 2.0 kg * m/s (B) 9.0 kg * m/s (C) 13 kg * m/s (D) 25 kg * m/s

(B) 9.0 kg * m/s

(A) 2RT - S (B) 2RT / M - S / M (C) 1/3RT^3 - 1/2ST^2 + ZT (D) 1/3 RT^3/M - 1/2 ST^2/M + ZT/M (E) RT^2/M - ST/M + Z/M

(D) 1/3 RT^3/M - 1/2 ST^2/M + ZT/M

(A) m1 | v1f - v1i| (B) m1 (v1f + v1i) (C) m1 | v2f - v2i | (D) m1 (v2f + v2i)

(B) m1 (v1f + v1i)

(A) Zero (B) 0.50 kg * m/s (C) 7.5 kg * m/s (D) 30 kg * m/s

(B) 0.50 kg * m/s

(A) Both students have the same final speed because the 50 kg student-chair system had the 10kg ball-40kg student-chair system have the same mass and experience impulses of the same magnitude (B) Both students have the same final speed because each ball-char student system experiences the same impulse regardless of the mass of the system (C) The 40-kg student has a greater final speed because the 40-kg student-chair system has less mass than the 50-kg student-chair system (D) The 40-kg student has a greater final speed because the ball transfers all of its momentum to the student when it is caught

(C) The 40-kg student has a greater final speed because the 40-kg student-chair system has less mass than the 50-kg student-chair system

(A) The slope of the graph line would be less. (B) The slope of the graph line would be greater. (C) The graph line would no longer be linear. (D) There would be no effect on the graph. (E) The change cannot be determined without knowing the ratio of m1/m2

(B) The slope of the graph line would be greater.

An object is moving in a straight line down along the x-axis. The net force Fx exerted on the object as a function of time t is shown in the graph. Which of the following coul be momentum px of the object as a function of t? (A) (B) (C) (D) (E)

(E)

(A) 30 s (B) 60 s (C) 120 s (D) 180 s (E) 240 s

(D) 180 s

(A) v (B) 2v (C) 3v (D) 4v (E) 5v

(A) The average force exerted by the floor (B) The amount of time in contact with the floor (C) The impulse exerted by the floor (D) The momentum just before colliding with the floor (E) The kinetic energy just before colliding with the floor

A truck of mass mT is initially at rest with a load of sand of initial mass ms. At time t = 0, the truck is pushed forward by a constant force F0 and the sand begins to leave the truck, as shown. The mass of the sand that has left the truck as a function of time is given by Ct, where C is positive constant with appropriate units. At time t1 there is no sand left in the truck. Which of the following graphs most nearly shows the velocity vT of the truck as a function of time? (A) (B) (C) (D)

(C) The velocity of the truck will change at an increasing rate during the time interval 0 < t < t1 because the mass of the truck is contents is decreasing over time. .After t1 the rate at which the velocity of the truck changes is constant because the net force of the truck and the mass of the truck are both constant.

(A) The net force is decreasing over time. (B) The net force is increasing over time. (C) The net force is constant and positive. (D) The net force is constant and negative. (E) The net force is positive for 0 < t < 6 s and negative for 6s < t < 10 s.

(D) The net force is constant and negative.

(A) 4v(t-t1) (B) D + 4v(t-t1) (C) D + 2/3v(t-t1) (D) D + 4/3v(t-t1)

(D) D + 4/3v(t-t1)

(A) v0(1-cos(theta)) and -v0sin(theta) (B) -v0(cos(theta)) and -v0sin(theta) (C) v0 and -v0 (D) 0 and 0

(A) v0(1-cos(theta)) and -v0sin(theta)

(A) The momentum of the two-object system is constant, and the magnitude of the relative velocity between the objects is the same before and after the collision. (B) The momentum of the two-object system is constant, and the magnitude of the relative velocity between the objects is different before and after the collision. (C) The momentum of the 1-kg object is constant, and the magnitude of the relative velocity between the objects is the same before and after the collision. (D) The momentum of the 1-kg object is not constant, and the magnitude of the relative velocity between the objects is different before and after the collision.

(A) The momentum of the two-object system is constant, and the magnitude of the relative velocity between the objects is the same before and after the collision.

(A) Car 1 was more massive than car 2 (B) Car 2 was more massive than car 1 (C) The velocity measurements for car 1 were lower than the car's actual speed (D) The velocity measurements for car 2 were higher than the car's actual speed (E) The experiment was performed without error

(B) Car 2 was more massive than car 1

(A) They must be moving to the left. (B) They must be moving to the right. (C) They must be at rest. (D) The combined objects' direction of motion, if any, cannot be determined without knowing the ratio of the speeds of the objects before the collision

(D) The combined objects' direction of motion, if any, cannot be determined without knowing the ratio of the speeds of the objects before the collision

(A) The slope of the line would be less. (B) The slope of the line would be greater. (C) The line would become a curve with a slope that is not constant. (D) The line would not change

(B) The slope of the line would be greater.

(A) Increases and Decreases (B) Increases and Stays the same (C) Stays the same and Increases (D) Stays the same and Stays the same

(B) Increases and Stays the same

(A) vPcm = vQcm = vRcm (B) vQcm < vRcm < vPcm (C) vRcm < vPcm < vQcm (D) vPcm < vQcm < vRcm

(A) vPcm = vQcm = vRcm

(A) v0 (B) 2v0 / 3 (C) v0 / 2 (D) v0 / 3

(D) v0 / 3

(A) v1 = v2 and J1 > J2 (B) v1 = v2 and J 1 = J2 (C) v1 > v2 and J1 > J2 (D) v1 > v2 and J1 = J2

(A) v1 = v2 and J1 > J2

(A) m2 (v2f - v2i) (B) m1 (v2f - v2i) (C) m2 (v1f - v1i) (D) (2m1v1i) / (m1 + m2) (E) (m1-m2)v1i / (m1 + m2)

(A) m2 (v2f - v2i)

(A) Vf2 > 2Vf (B) Vf < Vf2 < 2Vf (C) Vf2 = Vf (D) 1/2Vf < Vf2 < Vf (E) Vf2 < 1/2Vf

(A) The forces the truck and car exert on each other must be external to the truck-car system because the momentum of the truck changes. (B) The forces the truck and car exert on each other must be external to the truck-car system because the momentum of the car changes. (C) The forces the truck and car exert on each other must be external to the truck-car system because the momentum of both the truck and car change. (D) The forces the truck and car exert on each other must be internal to the truck-car system because the momentum of the center of mass of the truck-car system stays the same. (E) The forces the truck and car exert on each other must be internal to the truck-car system because the momentum of the center of mass of the truck-car system changes

(D) The forces the truck and car exert on each other must be internal to the truck-car system because the momentum of the center of mass of the truck-car system stays the same

(A) 1/5 V (B) 4/5 V (C) V (D) 5/4 V (E) 5 V

(B) 4/5 V

(A) 1 kg*m/s (B) 2 kg * m/s (C) 4 kg * m/s (D) 6 kg * m/s (E) 8 kg * m/s

(D) 6 kg * m/s

(A) 4.0 N * s (B) 8.5 N * s (C) 13 N * s (D) 25 N * s (E) 30 N * s

(B) 8.5 N * s

(A) 40 kg * m/s (B) 20 kg * m/s (C) 0 kg * m/s (D) -20 kg * m/s (E) indeterminable unless the mass M of the object is known

(A) 6 kg * m/s (B) 10 kg * m/s (C) 12 kg * m/s (D) 30 kg * m/s (E) It cannot be determined without knowing the initial momentum of the object

(A) 0 (B) 0.67 kg * m/s (C) 3.33 kg * m/s (D) 3.60 kg * m/s (E) The change in momentum of the ball cannot be determined without knowing the mass of the ball

(B) 0.67 kg * m/s

(A) 2Ft (B) Ft (C) 1/2 Ft (D) 1/4 Ft (E) zero

(A) m1 + m2 + 3m3 (B) m1 + m2 = m3 (C) m1 + m2 + m3 = 0 (D) m1 + m2 = 1/2 m3

(A) m1 + m2 + 3m3

(A) zero (B) -v/2 (C) -2v/3 (D) -3v/2 (E) -2v

(A) zero

A) Scenario 1 (B) Scenario 2 (C) Scenario 3 (D) The speed of the center of mass is the same in all three scenarios

(A) v1 = 1/3 v and v2 = 2/3 v (B) v1 = -1/3 v and v2 = 2/3 v (C) v1 = 0 and v2 = 1/2v (D) v1 = -1/2 v and v2 = 1/2 v (E) v1 = 1/2 v and v2 = 1/2 v

(B) v1 = -1/3 v and v2 = 2/3 v

(A) Greater-than-expected friction between the block and the surface (B) Deformation of the block as a result of the projectile’s impact (C) Friction between the projectile and the block as the projectile becomes embedded in the block (D) An increase in temperature of the block as a result of the impact (E) A slight downward incline of the surface the block moves on

(E) A slight downward incline of the surface the block moves on

(A) (m1vi) / 2(△t) (B) (m2vi) / 2(△t) (C) (m1Vf) / m2(△t) (D) (m1vf) / △t (E) (m2vf) / △t

(E) (m2vf) / △t

(A) 2v0/3 to the right (B) v0/3 to the left (C) v0/3 to the right (D) 2v0/3 to the right (E) 5v0/3 to the left

(B) v0/3 to the left

(A) 2/3 m/s (B) 1 m/s (C) 5/3 m/s (D) 10/3 m/s (E) 7m/s

(A) 2/3 m/s

(A) bt/m (B) bt^2/2m (C) bsqrt(t) / m (D) b/mt

(B) bt^2/2m

(A) 22 degrees (B) 36 degrees (C) 45 degrees (D) 54 degrees (E) 62 degrees

(A) 22 degrees

(A) Zero (B) mg (C) 2mg (D) 4mg (E) It cannot be determined without knowing the length of time of that the ball is in contact with the floor

(E) It cannot be determined without knowing the length of time of that the ball is in contact with the floor

(A) 2m(vf - vi) (B) m(vf-vi) (C) m(vf + vi) (D) mvi (E) mvf

(A) 0.5 m/s (B) 0.8 m/s (C) 2.0 m/s (D) 4.0 m/s (E) 8.0 m/s

(B) 0.8 m/s

(A) Same direction as Pucks A and B / The initial momentum of Puck C is the same as the initial momentum of Puck A (B) Same direction as Pucks A and B / Both collisions are inelastic (C) Different direction than Pucks A and B / The speed of Puck C is greater than the speed of Puck A (D) Different direction than Pucks A and B / The total mass of the two-puck systems is different in each collision

(A) Same direction as Pucks A and B / The initial momentum of Puck C is the same as the initial momentum of Puck A

(A) 1/2vB (B) 3/4vB (C) vB (D) 3/2vB

(D) 3/2vB

Unit 4 Test Flashcards

(100 cards)