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An object of mass 2 kg collides with an object of mass 1 kg that is at rest, as shown in the figure. A graph of the force as a function of time that the 1 kg object exerts on the 2 kg object is shown. After the collision, the 2 kg object has a speed of 2 , and the 1 kg object has a speed of 8 . A student must determine the initial speed vo of the 2 kg object. Which of the following options shows the correct substitutions into the equation Ap = FAt so that the initial speed of the unknown object can be determined? (2 kg)(2 - vo) = (8000 N)(0.00125 s), because the maximum force exerted on the 2 kg object must be considered. (2 kg)(2 – vo) = ÷(8000 N)(0.00125 s), because the average force exerted on the 2 kg object must be considered. (2 kg – 1 kg)(2 -vo) = (8000 N)(0.00125 s), because the difference in mass and the maximum force exerted on the 2 kg object must be considered. (2 kg – 1 kg)(2 m/s – vo) = (8000 N)(0.00125 s), because the difference in mass and the average force exerted on the 2 kg object must be considered.

An object of mass 2 kg collides with an object of mass 1 kg that is at rest, as shown in the figure. A graph of the force as a function of time that the 1 kg object exerts on the 2 kg object is shown. After the collision, the 2 kg object has a speed of 2 , and the 1 kg object has a speed of 8 . A student must determine the initial speed vo of the 2 kg object. Which of the following options shows the correct substitutions into the equation Ap = FAt so that the initial speed of the unknown object can be determined? (2 kg)(2 - vo) = (8000 N)(0.00125 s), because the maximum force exerted on the 2 kg object must be considered. (2 kg)(2 – vo) = ÷(8000 N)(0.00125 s), because the average force exerted on the 2 kg object must be considered. (2 kg – 1 kg)(2 -vo) = (8000 N)(0.00125 s), because the difference in mass and the maximum force exerted on the 2 kg object must be considered. (2 kg – 1 kg)(2 m/s – vo) = (8000 N)(0.00125 s), because the difference in mass and the average force exerted on the 2 kg object must be considered.

Astronomy
Astronomy
1st Edition
ISBN:9781938168284
Author:Andrew Fraknoi; David Morrison; Sidney C. Wolff
Publisher:Andrew Fraknoi; David Morrison; Sidney C. Wolff
Chapter19: Celestial Distances
Section: Chapter Questions
Problem 7E: While a meter is the fundamental unit of length, most distances traveled by humans are measured in...
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An object of mass 2 kg collides with an object of mass 1 kg that is at rest, as shown in the figure. A graph of the force as a function of time that the 1 kg object exerts on the 2 kg object is shown. After the collision, the 2 kg object has a speed of 2 m, and the 1 kg object has a speed of 8 m. A student must determine the initial speed vo of the 2 kg object. Which of the following options shows the correct substitutions into the equation Ap = FAt so that the initial speed of the unknown object can be determined? (2 kg)(2 m – vo) = (8000 N)(0.00125 s), because the maximum force exerted on the 2 kg object must be considered. (2 kg)(2 - vo) (8000 N)(0.00125 s), because the average force exerted on m S the 2 kg object must be considered. (2 kg – 1 kg)(2 -vo) (8000 N)(0.00125 s), because the difference in mass and the maximum force exerted on the 2 kg object must be considered. (2 kg – 1 kg)(2 m/s – vo) = ;(8000 N)(0.00125 s), because the difference in mass and the average force exerted on the 2 kg object must be considered.
Transcribed Image Text:An object of mass 2 kg collides with an object of mass 1 kg that is at rest, as shown in the figure. A graph of the force as a function of time that the 1 kg object exerts on the 2 kg object is shown. After the collision, the 2 kg object has a speed of 2 m, and the 1 kg object has a speed of 8 m. A student must determine the initial speed vo of the 2 kg object. Which of the following options shows the correct substitutions into the equation Ap = FAt so that the initial speed of the unknown object can be determined? (2 kg)(2 m – vo) = (8000 N)(0.00125 s), because the maximum force exerted on the 2 kg object must be considered. (2 kg)(2 - vo) (8000 N)(0.00125 s), because the average force exerted on m S the 2 kg object must be considered. (2 kg – 1 kg)(2 -vo) (8000 N)(0.00125 s), because the difference in mass and the maximum force exerted on the 2 kg object must be considered. (2 kg – 1 kg)(2 m/s – vo) = ;(8000 N)(0.00125 s), because the difference in mass and the average force exerted on the 2 kg object must be considered.
+y >+X 2 kg 1 kg 10,000 8000 2 6000 4000 2000 0 - 0.0010 0.0015 0.0020 0.0025 t (s)
Transcribed Image Text:+y >+X 2 kg 1 kg 10,000 8000 2 6000 4000 2000 0 - 0.0010 0.0015 0.0020 0.0025 t (s)
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