Part 9 of 10-Analyze Now Jed and Kadia tackle a homework problem: An object of mass m, = 13 kg and velocity v, = 5.5 m/s crashes into another object of mass m, = 6 kg and velocity v2 = -20.5 m/s. The two particles stick together as a result of the collision. Because no external forces are acting, the collision does not change the total momentum of the system of two particles, so the principle of conservation of linear momentum applies. m,V1i + m2V2i = (m, + m2)Vf %3D If Jed and Kadia use the one-dimensional conservation of momentum equation to find the final velocity of the two joined objects after the collision, what do they obtain? (Indicate the direction with the sign of your answer.) -2.71 -2.71 m/s Part 10 of 10 - Analyze Although the forces that the two objects exert on each other cannot change their total momentum, they can change the total kinetic energy in an inelastic collision such as the one being considered. All of the mechanical energy in the problem is kinetic energy. Change in kinetic energy is given by AK = What.do Jed.and Kadia find for the change in the total mechanical energy of the system?

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Chapter1: Units, Trigonometry. And Vectors
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**Part 9 of 10 - Analyze**

Now Jed and Kadia tackle a homework problem:

An object of mass \(m_1 = 13 \, \text{kg}\) and velocity \(\vec{v}_1 = 5.5 \, \text{m/s}\) crashes into another object of mass \(m_2 = 6 \, \text{kg}\) and velocity \(\vec{v}_2 = -20.5 \, \text{m/s}\). The two particles stick together as a result of the collision. Because no external forces are acting, the collision does not change the total momentum of the system of two particles, so the principle of conservation of linear momentum applies.

\[
m_1 \vec{v}_1 + m_2 \vec{v}_2 = (m_1 + m_2) \vec{v}_f
\]

If Jed and Kadia use the one-dimensional conservation of momentum equation to find the final velocity of the two joined objects after the collision, what do they obtain? (Indicate the direction with the sign of your answer.)

-2.71 ✔️ -2.71 m/s


**Part 10 of 10 - Analyze**

Although the forces that the two objects exert on each other cannot change their total momentum, they can change the total kinetic energy in an inelastic collision such as the one being considered. All of the mechanical energy in the problem is kinetic energy. Change in kinetic energy is given by

\[
\Delta K = \frac{1}{2} (m_1 + m_2) \vec{v}_f^2 - \left( \frac{1}{2} m_1 \vec{v}_1^2 + \frac{1}{2} m_2 \vec{v}_2^2 \right)
\]

What do Jed and Kadia find for the change in the total mechanical energy of the system?

*Enter a number.*

[Submit]
Transcribed Image Text:**Part 9 of 10 - Analyze** Now Jed and Kadia tackle a homework problem: An object of mass \(m_1 = 13 \, \text{kg}\) and velocity \(\vec{v}_1 = 5.5 \, \text{m/s}\) crashes into another object of mass \(m_2 = 6 \, \text{kg}\) and velocity \(\vec{v}_2 = -20.5 \, \text{m/s}\). The two particles stick together as a result of the collision. Because no external forces are acting, the collision does not change the total momentum of the system of two particles, so the principle of conservation of linear momentum applies. \[ m_1 \vec{v}_1 + m_2 \vec{v}_2 = (m_1 + m_2) \vec{v}_f \] If Jed and Kadia use the one-dimensional conservation of momentum equation to find the final velocity of the two joined objects after the collision, what do they obtain? (Indicate the direction with the sign of your answer.) -2.71 ✔️ -2.71 m/s **Part 10 of 10 - Analyze** Although the forces that the two objects exert on each other cannot change their total momentum, they can change the total kinetic energy in an inelastic collision such as the one being considered. All of the mechanical energy in the problem is kinetic energy. Change in kinetic energy is given by \[ \Delta K = \frac{1}{2} (m_1 + m_2) \vec{v}_f^2 - \left( \frac{1}{2} m_1 \vec{v}_1^2 + \frac{1}{2} m_2 \vec{v}_2^2 \right) \] What do Jed and Kadia find for the change in the total mechanical energy of the system? *Enter a number.* [Submit]
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