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the graph above shows the position x as a function of time for the center of mass of a system of particles of total mass 6.0 kg. for a very short time interval around 2.0 s, an external force is exerted on an object in the system. what is the resulting change in momentum of the system?

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Answer:

To find the resulting change in momentum of the system, we need to calculate the momentum of the system before and after the external force is exerted, and then find the difference.

The momentum of a system of particles is the product of the total mass of the system and the velocity of its center of mass. From the graph, we can see that the velocity of the center of mass at t = 2.0 s is approximately 0.8 m/s.

Before the external force is exerted, the momentum of the system is:

p1 = m*v1 = 6.0 kg * 0.8 m/s = 4.8 kg m/s

After the external force is exerted, the velocity of the center of mass changes, and we can estimate it from the graph to be approximately -0.4 m/s at t = 2.0 s + Δt, where Δt is a very short time interval. The momentum of the system after the external force is exerted is:

p2 = m*v2 = 6.0 kg * (-0.4 m/s) = -2.4 kg m/s

The resulting change in momentum of the system is:

Δp = p2 - p1 = (-2.4 kg m/s) - (4.8 kg m/s) = -7.2 kg m/s

Therefore, the resulting change in momentum of the system is -7.2 kg m/s.

User Mohammad Olfatmiri
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