# Physics Honors Test Prep

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Test Prep for Test 2 Physics Honors at Pine Crest School.

• 1.

### Suppose a 60-kg boy and a 41-kg girl use a massless rope in a tug-of-war on an icy, resistance-free surface. If the acceleration of the girl toward the boy is 3.0 m/s2, find the magnitude of the acceleration of the boy toward the girl.

• A.

2.05 m/s2

• B.

2 m/s2

• C.

2.1 m/s2

A. 2.05 m/s2
Explanation
The magnitude of the acceleration of the boy toward the girl can be found using Newton's second law of motion, which states that force is equal to mass multiplied by acceleration. Since the girl's acceleration is given as 3.0 m/s^2 and the girl's mass is 41 kg, the force exerted by the girl is 3.0 * 41 = 123 N. Since the rope is massless, the force exerted by the boy is also 123 N. Using the boy's mass of 60 kg, we can calculate the boy's acceleration using the formula force = mass * acceleration. Rearranging the formula, we get acceleration = force / mass, which gives us 123 / 60 = 2.05 m/s^2. Therefore, the magnitude of the acceleration of the boy toward the girl is 2.05 m/s^2.

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• 2.

### As a baseball is being caught, its speed goes from 29.0 m/s to 0.0 m/s in about 0.0050 s. The mass of the baseball is 0.130 kg. (a) What is the baseball's acceleration?

• A.

15800 m/s2

• B.

16000 m/s2

• C.

15800.0 m/s2

• D.

-16000 m/s2

A. 15800 m/s2
Explanation
The acceleration of an object can be calculated using the formula: acceleration = (final velocity - initial velocity) / time. In this case, the final velocity is 0.0 m/s, the initial velocity is 29.0 m/s, and the time is 0.0050 s. Plugging these values into the formula, we get: acceleration = (0.0 m/s - 29.0 m/s) / 0.0050 s = -29.0 m/s / 0.0050 s = -5800 m/s^2. However, since acceleration is a vector quantity, it has both magnitude and direction. The negative sign indicates that the acceleration is in the opposite direction of the initial velocity. Therefore, the correct answer is -5800 m/s^2 or -16000 m/s^2.

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• 3.

### As a baseball is being caught, its speed goes from 29.0 m/s to 0.0 m/s in about 0.0050 s. The mass of the baseball is 0.130 kg. (b) What are the magnitude and direction of the force acting on it? *Pick 2 answers*

• A.

754 N

• B.

750 N

• C.

Same direction as velocity of the ball

• D.

Opposite in direction to velocity of the ball

A. 754 N
D. Opposite in direction to velocity of the ball
Explanation
The magnitude of the force acting on the baseball can be calculated using Newton's second law, which states that force is equal to mass multiplied by acceleration. In this case, the acceleration can be determined by dividing the change in velocity (29.0 m/s - 0.0 m/s) by the time it takes for the velocity to change (0.0050 s). The direction of the force is opposite to the direction of the velocity, as the baseball is being caught and its velocity is decreasing. Therefore, the correct answer is 754 N, opposite in direction to the velocity of the ball.

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• 4.

### As a baseball is being caught, its speed goes from 29.0 m/s to 0.0 m/s in about 0.0050 s. The mass of the baseball is 0.130 kg. (c) What are the magnitude and direction of the force acting on the player who caught it? *Pick 2 answers*

• A.

754 N

• B.

750 N

• C.

Same direction as velocity of the ball

• D.

Opposite in direction to the velocity of the ball

A. 754 N
C. Same direction as velocity of the ball
Explanation
The magnitude of the force acting on the player who caught the baseball can be calculated using Newton's second law, which states that force equals mass times acceleration. In this case, the acceleration can be calculated as the change in velocity divided by the time taken. The change in velocity is 29.0 m/s (initial velocity) minus 0.0 m/s (final velocity), and the time taken is 0.0050 s. Plugging these values into the equation, we get force = (0.130 kg) * ((29.0 m/s - 0.0 m/s) / 0.0050 s) = 754 N. The force is in the same direction as the velocity of the ball because the player is exerting a force to stop the ball's motion.

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• 5.

### A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 844 N. (a) As the elevator moves up, the scale reading increases to 930 N. Find the acceleration of the elevator.

• A.

1 m/s2

• B.

0.999 m/s2

• C.

0.99 m/s2

B. 0.999 m/s2
Explanation
When the elevator is at rest, the scale reading is equal to the weight of the student, which is 844 N. As the elevator moves up, the scale reading increases to 930 N. This increase in the scale reading is due to the additional force exerted on the student by the elevator's acceleration. Using Newton's second law, we can calculate the net force on the student by subtracting the weight from the scale reading. The net force is equal to the mass of the student multiplied by the acceleration of the elevator. Rearranging the equation, we can solve for the acceleration, which is approximately 0.999 m/s2.

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• 6.

### A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 844 N. (b) As the elevator approaches the 74th floor, the scale reading drops to 782 N. What is the acceleration of the elevator?

• A.

-0.720 m/s2

• B.

0.720 m/s2

• C.

-1 m/s2

• D.

1 m/s2

A. -0.720 m/s2
Explanation
The scale reading on the bathroom scale decreases as the elevator approaches the 74th floor. This indicates that there is a decrease in the normal force acting on the student. According to Newton's second law, the net force acting on an object is equal to its mass multiplied by its acceleration. Since the mass of the student remains constant, the decrease in the normal force must be due to a decrease in the net force. This decrease in net force can be explained by a downward acceleration of the elevator. Therefore, the acceleration of the elevator is -0.720 m/s2.

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• 7.

### (c) Using your results from parts a and b, select which change in velocity, starting or stopping, takes the longer time. Think of an explanation.

• A.

Stopping

• B.

Starting

• C.

Neither

A. Stopping
Explanation
Based on the results from parts a and b, it can be concluded that stopping takes longer than starting. This is because when an object is stopping, it needs to overcome its initial velocity and decelerate to a complete stop. On the other hand, when an object is starting, it only needs to accelerate from rest to its final velocity. Therefore, the process of stopping requires more time and distance to be covered compared to starting.

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• Sep 29, 2009
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