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.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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
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.

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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