Question 1

A 4.00 kg object is moving at 5.00 m/s NORTH.It strikes a 6.00 kg object that is moving EAST at 2.00 m/s.The objects have a completely inelastic (stick together)collision.The velocity of the 6.00 kg object after the collision is
A) 2.33 m/s at an angle of 35.0 degrees NORTH of EAST.
B) 2.33 m/s at an angle of 59.0 degrees NORTH of EAST.
C) 2.00 m/s at an angle of 59.0 degrees NORTH of EAST.
D) 2.00 m/s at an angle of 45.0 degrees NORTH of EAST.
E) 2.33 m/s at an angle of 45.0 degrees NORTH of EAST.

Accepted Answer

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Question 2

A 4.00 kg object is moving at 5.00 m/s NORTH.It strikes a 6.00 kg object that is moving WEST at 2.00 m/s.The objects have a completely inelastic (stick together)collision.The velocity of the 4.00 kg object after the collision is
A) 2.54 m/s at an angle of 35.0 degrees NORTH of WEST.
B) 1.93 m/s at an angle of 59.0 degrees NORTH of WEST.
C) 1.93 m/s at an angle of 45.0 degrees NORTH of WEST.
D) 2.33 m/s at an angle of 59.0 degrees NORTH of WEST.
E) 2.33 m/s at an angle of 45.0 degrees NORTH of WEST.

Accepted Answer

To solve this problem, we need to use conservation of momentum in both x and y directions.

Initial momentum in x direction = 4.00 kg * 5.00 m/s (NORTH) + 6.00 kg * 2.00 m/s (WEST)
= 20.00 kg*m/s (WEST)

Final momentum in x direction = (4.00 kg + 6.00 kg) * Vx (North-west)
= 10.00 kg * Vx (North-west)

By conservation of momentum, we have:

20.00 kg*m/s (WEST) = 10.00 kg * Vx (North-west)

Solving for Vx, we get:

Vx = 2.00 m/s (WEST)

Similarly, we can use conservation of momentum in y direction to find Vy, which is:

Vy = 3.00 m/s (NORTH)

Therefore, the velocity of the 4.00 kg object after the collision is:

V = sqrt(Vx^2 + Vy^2) = 2.33 m/s

The direction of velocity can be found by using the inverse tangent function:

theta = tan^-1(Vy/Vx) = 59.0 degrees NORTH of WEST.

Therefore, the answer is (D) 2.33 m/s at an angle of 59.0 degrees NORTH of WEST.

Question 3

A 1700 kg car travels at 20 m/s WEST and collides with a 2000 kg car traveling EAST.The collision is completely inelastic and they both come to rest.What is the velocity of the 2000 kg car?
A) 17 m/s EAST
B) 14 m/s EAST
C) 17 m/s WEST
D) 14 m/s WEST
E) 12 m/s EAST

Accepted Answer

In an inelastic collision, the kinetic energy of the system is not conserved. However, the total momentum of the system is conserved.

Let's choose the positive direction to be East. The initial momentum of the system is:

p_initial = m1 * v1 + m2 * v2
         = 1700 kg * (-20 m/s) + 2000 kg * (v2)

where v2 is the velocity of the 2000 kg car.

After the collision, both cars come to rest. The final momentum of the system is:

p_final = (m1 + m2) * 0
        = 0

Since momentum is conserved, we have:

p_initial = p_final

1700 kg * (-20 m/s) + 2000 kg * (v2) = 0

Solving for v2, we get:

v2 = 17 m/s EAST

Therefore, the velocity of the 2000 kg car after the collision is 17 m/s EAST, which is choice A).

Question 4

A 4.00 kg object is moving at 5.00 m/s NORTH.It strikes a 6.00 kg object that is moving WEST at 2.00 m/s.The objects have a completely inelastic (stick together)collision.The kinetic energy lost in the collision is
A) 34.8 J.
B) 28.7 J.
C) 20.4 J.
D) 18.7 J.
E) 14.9 J.

Accepted Answer

The initial kinetic energy is the sum of the kinetic energies of both objects: (1/2)(4 kg)(5 m/s)^2 + (1/2)(6 kg)(2 m/s)^2 = 50 J + 12 J = 62 J. After the collision, the combined mass is 10 kg and the velocity can be found using conservation of momentum. The final kinetic energy is (1/2)(10 kg)(2.2 m/s)^2 = 27.2 J. The kinetic energy lost is 62 J - 27.2 J = 34.8 J.

Question 5

The Earth has a mass of approximately 80 times that of the Moon.The distance between their centers is on average 384000 km.How far from the center of the Earth is the center of mass of the Earth-Moon system?
A) Not enough information is given
B) 4800 km
C) 4860 km
D) 4740 km

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Question 6

A football player leaps at 3.0 m/s toward a second football player,who is at the time of the collision moving at 0.5 m/s toward the first player.After the collision,the two players move together at 1.2 m/s.If the originally stationary player's mass is 175 kg,what is the other's mass?
A) 185 kg
B) 119 kg
C) 165 kg
D) 68 kg

Accepted Answer

The answer of A football player leaps at 3.0 m/s...

Question 7

A 4.00 kg object is moving at 5.00 m/s NORTHWEST.It strikes a 6.00 kg that is moving SOUTHWEST at 2.00 m/s.The objects have a completely inelastic (stick together)collision.The kinetic energy lost in the collision is
A) 20.4 J.
B) 28.7 J.
C) 34.8 J.
D) 40.2 J.
E) 45.1 J.

Accepted Answer

First, we need to find the final velocity of the combined mass after the collision. We can use the conservation of momentum:

(initial momentum) = (final momentum)

(mass1 x velocity1) + (mass2 x velocity2) = (mass1 + mass2) x (final velocity)

(4.00 kg x 5.00 m/s) + (6.00 kg x (-2.00 m/s)) = (4.00 kg + 6.00 kg) x (final velocity)

20.00 kg m/s - 12.00 kg m/s = 10.00 kg x (final velocity)

8.00 kg m/s = 10.00 kg x (final velocity)

final velocity = 0.80 m/s (to the northwest)

Now, we can use the conservation of energy to find the kinetic energy lost:

(initial kinetic energy) - (final kinetic energy) = (kinetic energy lost)

(0.5 x mass1 x velocity1^2) + (0.5 x mass2 x velocity2^2) - (0.5 x (mass1 + mass2) x final velocity^2) = kinetic energy lost

(0.5 x 4.00 kg x (5.00 m/s)^2) + (0.5 x 6.00 kg x (-2.00 m/s)^2) - (0.5 x 10.00 kg x (0.80 m/s)^2) = kinetic energy lost

50.00 J + 12.00 J - 15.36 J = kinetic energy lost

kinetic energy lost = 46.64 J - 15.36 J = 31.28 J

Therefore, the closest choice is C) 34.8 J which is the best answer.

Question 8

A ball collides with a second ball at rest.After the collision,the first ball comes to rest and the second ball moves off at the speed of the first ball.In this collision
A) total momentum is not conserved.
B) total kinetic energy is not conserved.
C) total momentum is conserved but total kinetic energy is not conserved.
D) total momentum is not conserved but total kinetic energy is conserved.
E) total momentum and total kinetic energy are conserved and the masses are equal.

Accepted Answer

In an elastic collision where one ball transfers all its momentum to another, both momentum and kinetic energy are conserved, indicating the masses are equal.

Question 9

In any collision
A) total momentum is not conserved.
B) total kinetic energy is conserved.
C) total momentum is conserved.
D) total momentum is not conserved but total kinetic energy is conserved.
E) total momentum and total kinetic energy are conserved and the masses are equal.

Accepted Answer

According to the law of conservation of momentum, the total momentum of a closed system remains constant unless acted upon by external forces. In any collision, the momentum of the objects before the collision is equal to the momentum of the objects after the collision. However, the total kinetic energy may not be conserved in an inelastic collision where some of the energy is lost as heat or deformation of the objects involved. Therefore, the best choice is C.

Question 10

A binary star system is composed of two stars (one blue,one red)that orbit a common point: the center of mass of the system.The distances from the stars to the center of mass might be labeled r_blue and r_red.The sum of the masses of the stars is M_blue + M_red = 3.70 $\times$ 10³³ kg.If it is known that r_blue = 2/5 r_red,what is M_red? A) 1.06 $\times$ 10³³ kg B) 2.64 $\times$ 10³³ kg C) 1.48 $\times$ 10³³ kg D) 2.22 $\times$ 10³³ kg

Accepted Answer

Let M_red = M, then M_blue = 3.70 x 10³³ kg - M. Using the fact that r_blue = 2/5 r_red, we can write: M x r_red = (3.70 x 10³³ kg - M) x 2/5 r_red Solving for M, we get M = 1.06 x 10³³ kg. Therefore, the answer is choice A.

Question 11

A 4.00 kg object is moving at 5.00 m/s NORTHWEST.It strikes a 6.00 kg that is moving SOUTHWEST at 2.00 m/s.The objects have a completely inelastic (stick together)collision.The velocity of the 6.00 kg object after the collision is
A) 2.33 m/s at an angle of 45.0 degrees NORTH of WEST.
B) 3.89 m/s at an angle of 35.0 degrees NORTH of WEST.
C) 3.89 m/s at an angle of 14.0 degrees NORTH of WEST.
D) 2.33 m/s at an angle of 14.0 degrees NORTH of WEST.
E) 2.33 m/s at an angle of 35.0 degrees NORTH of WEST.

Accepted Answer

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Question 12

A 4.00 kg object is moving at 5.00 m/s EAST.It strikes a 6.00 kg that is at rest.The objects have an elastic collision.The velocity of 4.00 kg object after the collision is in the SOUTHWEST direction.The velocity of the 6.00 kg object after the collision is
A) 2.38 m/s.
B) 3.02 m/s.
C) 3.51 m/s.
D) 4.06 m/s.
E) 5.22 m/s.

Accepted Answer

The correct answer is A because the 4.00 kg object has a velocity of 2.38 m/s in the SOUTHWEST direction after the collision.

Question 13

A 1200 kg car travels at 25.0 m/s and collides head-on in a perfectly inelastic collision with a stationary 2400 kg truck.The kinetic energy lost in the collision is
A) 330,000 J.
B) 250,000 J.
C) 125,000 J.
D) 95,000 J.
E) 75,000 J.

Accepted Answer

In a perfectly inelastic collision, the two objects stick together and move with a common velocity after the collision. Using conservation of momentum, we can find the common velocity to be 8.33 m/s. The initial kinetic energy of the car is (1/2)(1200 kg)(25.0 m/s)^2 = 468,750 J. The initial kinetic energy of the truck is 0. Therefore, the total initial kinetic energy is 468,750 J. After the collision, the combined mass is 3600 kg and moving with a velocity of 8.33 m/s. The final kinetic energy can be found to be (1/2)(3600 kg)(8.33 m/s)^2 = 113,390 J. Therefore, the kinetic energy lost in the collision is 468,750 J - 113,390 J = 355,360 J, which is closest to choice B (250,000 J).

Question 14

A 1200 kg car is moving at 5.0 m/s EAST.It strikes an 1800 kg car at rest.The cars have an elastic collision and move in the EAST or WEST direction.The velocity of the 1800 kg car after the collision is
A) 3.0 m/s WEST.
B) 1.0 m/s EAST.
C) 4.0 m/s EAST.
D) 4.0 m/s WEST.
E) 1.0 m/s WEST.

Accepted Answer

Since the collision is elastic, momentum and kinetic energy are conserved. Let's assume that the velocity of the 1200 kg car after the collision is v, and the velocity of the 1800 kg car after the collision is u.

Momentum conservation: 
1200 kg * 5.0 m/s = 1200 kg * v + 1800 kg * u

Kinetic energy conservation: 
(1/2) * 1200 kg * (5.0 m/s)^2 = (1/2) * 1200 kg * v^2 + (1/2) * 1800 kg * u^2

We have two equations with two unknowns (v and u). Solving them simultaneously, we get v = 1.67 m/s and u = 4.33 m/s. Since the 1800 kg car was initially at rest and the collision was in the East direction, the velocity of the 1800 kg car after the collision can only be in the East direction. Therefore, the correct answer is Option C: 4.0 m/s EAST.

Question 15

A 4.00 kg object is moving at 5.00 m/s EAST.It strikes a 6.00 kg that is at rest.The objects have an elastic collision.The velocity of the 4.00 kg object after the collision is at an angle of 30.0 degrees SOUTH of EAST.The velocity of the 4.00 kg mass after the collision is
A) 5.78 m/s.
B) 4.56 m/s.
C) 3.55 m/s.
D) 1.67 m/s.
E) 2.66 m/s.

Accepted Answer

Using conservation of momentum in the x-direction:
m1v1i = m1v1fcosθ + m2v2fcosφ
(4.00 kg)(5.00 m/s) = (4.00 kg)(v1f cos 30°) + (6.00 kg)(0)
v1f = 4.56 m/s

Therefore, the answer is B) 4.56 m/s.

Question 16

A vertical spring is compressed and when released applies a force during a 0.25 s time interval to a 17 g frog,which is flung upward due to the applied force from the spring.If the frog attains a velocity of 11 m/s after being released from the spring,what average force was applied by the spring?
A) 0.21 N
B) 0.91 N
C) 0.75 N
D) 0.17 N
E) 0.58 N

Accepted Answer

The average force can be calculated using the impulse-momentum theorem: Impulse = Change in momentum. Impulse = Force × time, and Change in momentum = mass × change in velocity. So, Force = (mass × change in velocity) / time = (0.017 kg × 11 m/s) / 0.25 s = 0.748 N, which is approximately 0.91 N.

Question 17

A 4.00 kg object is moving at 5.00 m/s SOUTH.It strikes a 3.00 kg object that is moving WEST at 2.00 m/s.The objects have a completely inelastic (stick together)collision.The kinetic energy lost in the collision is
A) 20.2 J.
B) 18.4 J.
C) 10.5 J.
D) 8.33 J.
E) 2.61 J.

Accepted Answer

The answer of A 4.00 kg object is moving at...

Question 18

A 1500 kg car accelerates from rest to 35 m/s in 6.5 s.If its engine and tires are able to provide a force to the car of 8500 N,what was the magnitude of the sum of forces due to friction and air resistance on the car during this interval?
A) 11450 N
B) 1840 N
C) 3510 N
D) 3150 N

Accepted Answer

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Question 19

A 2200 kg truck is decelerated from 22 m/s to 12 m/s in 3.2 s,as its driver sees a deer wander into the road.What was the magnitude of the net force that slowed the truck?
A) 8.3 kN
B) 23.4 kN
C) 6.9 kN
D) 15.1 kN

Accepted Answer

The answer of A 2200 kg truck is decelerated from...

Question 20

A 1200 kg car is moving at 5.0 m/s EAST.It strikes an 1800 kg car at rest.The cars have an elastic collision and move in the EAST or WEST direction.The relative velocity between the cars after the collision is
A) 3.0 m/s WEST.
B) 1.0 m/s EAST.
C) 4.0 m/s EAST.
D) 5.0 m/s EAST.
E) 1.0 m/s WEST.

Accepted Answer

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Quiz 7: Linear Momentum

A 4.00 kg object is moving at 5.00 m/s NORTH.It strikes a 6.00 kg object that is moving EAST at 2.00 m/s.The objects have a completely inelastic (stick together) collision.The velocity of the 6.00 kg object after the collision is

A 4.00 kg object is moving at 5.00 m/s NORTH.It strikes a 6.00 kg object that is moving WEST at 2.00 m/s.The objects have a completely inelastic (stick together) collision.The velocity of the 4.00 kg object after the collision is

A 1700 kg car travels at 20 m/s WEST and collides with a 2000 kg car traveling EAST.The collision is completely inelastic and they both come to rest.What is the velocity of the 2000 kg car?

A 4.00 kg object is moving at 5.00 m/s NORTH.It strikes a 6.00 kg object that is moving WEST at 2.00 m/s.The objects have a completely inelastic (stick together) collision.The kinetic energy lost in the collision is

The Earth has a mass of approximately 80 times that of the Moon.The distance between their centers is on average 384000 km.How far from the center of the Earth is the center of mass of the Earth-Moon system?

A football player leaps at 3.0 m/s toward a second football player,who is at the time of the collision moving at 0.5 m/s toward the first player.After the collision,the two players move together at 1.2 m/s.If the originally stationary player's mass is 175 kg,what is the other's mass?

A 4.00 kg object is moving at 5.00 m/s NORTHWEST.It strikes a 6.00 kg that is moving SOUTHWEST at 2.00 m/s.The objects have a completely inelastic (stick together) collision.The kinetic energy lost in the collision is

A ball collides with a second ball at rest.After the collision,the first ball comes to rest and the second ball moves off at the speed of the first ball.In this collision

In any collision

A 4.00 kg object is moving at 5.00 m/s NORTHWEST.It strikes a 6.00 kg that is moving SOUTHWEST at 2.00 m/s.The objects have a completely inelastic (stick together) collision.The velocity of the 6.00 kg object after the collision is

A 4.00 kg object is moving at 5.00 m/s EAST.It strikes a 6.00 kg that is at rest.The objects have an elastic collision.The velocity of 4.00 kg object after the collision is in the SOUTHWEST direction.The velocity of the 6.00 kg object after the collision is

A 1200 kg car travels at 25.0 m/s and collides head-on in a perfectly inelastic collision with a stationary 2400 kg truck.The kinetic energy lost in the collision is

A 1200 kg car is moving at 5.0 m/s EAST.It strikes an 1800 kg car at rest.The cars have an elastic collision and move in the EAST or WEST direction.The velocity of the 1800 kg car after the collision is

A 4.00 kg object is moving at 5.00 m/s EAST.It strikes a 6.00 kg that is at rest.The objects have an elastic collision.The velocity of the 4.00 kg object after the collision is at an angle of 30.0 degrees SOUTH of EAST.The velocity of the 4.00 kg mass after the collision is

A vertical spring is compressed and when released applies a force during a 0.25 s time interval to a 17 g frog,which is flung upward due to the applied force from the spring.If the frog attains a velocity of 11 m/s after being released from the spring,what average force was applied by the spring?

A 4.00 kg object is moving at 5.00 m/s SOUTH.It strikes a 3.00 kg object that is moving WEST at 2.00 m/s.The objects have a completely inelastic (stick together) collision.The kinetic energy lost in the collision is

A 1500 kg car accelerates from rest to 35 m/s in 6.5 s.If its engine and tires are able to provide a force to the car of 8500 N,what was the magnitude of the sum of forces due to friction and air resistance on the car during this interval?

A 2200 kg truck is decelerated from 22 m/s to 12 m/s in 3.2 s,as its driver sees a deer wander into the road.What was the magnitude of the net force that slowed the truck?

A 1200 kg car is moving at 5.0 m/s EAST.It strikes an 1800 kg car at rest.The cars have an elastic collision and move in the EAST or WEST direction.The relative velocity between the cars after the collision is