91影视

The given data is listed below as,

• The mass of the object A is, $m_A=7\mathrm{kg}$ .
• The initial momentum of object A is, localid="1657860376383" ${\stackrel{鈫�}{P}}_{A,i}=(17,-5,0)kg.m/s$ .
• The mass of object B is, $m_B=11kg$.
• The initial momentum of object B is, ${\stackrel{鈫�}{P}}_{B,i}=(4,6,0)k.m/s$ .
• After the collision, the final momentum of object A is,${\stackrel{鈫�}{P}}_{A,f}=(13,3,0)kg.m/s.$ ${\stackrel{鈫�}{P}}_{A,f}=(13,3,0)kg.m/s$.

This law of conservation of momentum states that the total momentum of an object before and after collision with another object becomes equal.

The first law of thermodynamics states that the change in the internal energy of a system is equal to the difference between the heat added and the work done by the system.

The work-energy theorem states that the net change in the work is equal to the change in the kinetic energy.

The law of conservation of momentum gives the initial and the final momentum of objects A, B, and the system. The first law of thermodynamics and the work-energy theorem gives the increase in the internal energy of the system along with its kinetic energies of them.

The expression for the total initial momentum of the system is expressed as,

${\stackrel{鈫�}{P}}_{initial}={\stackrel{鈫�}{P}}_{A,i}+{\stackrel{鈫�}{P}}_{B,i}$

For${\stackrel{鈫�}{P}}_{A,i}=(17,-5,0)kg.m/s$ and${\stackrel{鈫�}{P}}_{B,i}=(4,6,0)kg.m/s$ ,

$$\begin{gathered} {\stackrel{鈫�}{P}}_{initial}=(17,-5,0)kg.m/s+(4,6,0)kg.m/s \\ =(21,1,0)kg.m/s \end{gathered}$$

Thus, the total initial momentum of the system is $(21,1,0)kg.m/s$.

The expression of the final momentum is expressed as,

${\stackrel{鈫�}{P}}_{initial}={\stackrel{鈫�}{P}}_{A,i}+{\stackrel{鈫�}{P}}_{B,i}$

As role="math" localid="1657861309005" ${\stackrel{鈫�}{P}}_{initial}={\stackrel{鈫�}{P}}_{initial}$ , then the above equation can be written as,

${\stackrel{鈫�}{P}}_{initial}={\stackrel{鈫�}{P}}_{A,i}+{\stackrel{鈫�}{P}}_{B,i}$

For${\stackrel{鈫�}{P}}_{initial}=(21,1,0)kg.m/s$and${\stackrel{鈫�}{P}}_{A,i}=(13,3,0)kg.m/s$,

$$\begin{gathered} (21,1,0)kg.m/s=(13,3,0)kg.m/s+{\stackrel{鈫�}{P}}_{B,i} \\ {\stackrel{鈫�}{P}}_{B,i}=(8,-2,0)kg.m/s \end{gathered}$$

Thus, the final momentum of object B is $(8,-2,0)kg.m/s$.

The equation of the initial kinetic energy of object A is expressed as,

$K.E_{.A,initial}=\frac{{\stackrel{鈫�}{P}}_{A,i^2}}{2m_A}$

Here, $p_{A,i}$ is the initial momentum of object A and $m_A$ is the mass of object A.

For${\stackrel{鈫�}{P}}_{A,i}=(17,-5,0)kg.m/s$and $m_A=7kg$,

$$\begin{gathered} K.E_{.A,initial}=\frac{{\left((17,-5,0)kg.m/s\right)}^2}{2脳7kg} \\ =\frac{\left((289,25,0)kg.m/s\right)}{14kg} \\ =(20.64,1.78,0)m/s \end{gathered}$$

Thus, the initial kinetic energy of object A is $(20.64,1.78,0)m/s$.

$p_{B,i}$The equation of the initial kinetic energy of the object B is expressed as,

localid="1657863580396" $K.E_{.B,initial}=\frac{{\stackrel{鈫�}{P}}_{B,i^2}}{2m_B}$

Here, $p_{B,i}$ is the initial momentum of object B and $m_B$ is the mass of object B.

For ${\stackrel{鈫�}{p}}_{B,i}=(4,6,0)kg.m/s$and $m_B=11kg$,

localid="1657863591107" $$\begin{gathered} K.E_{.B,initial}=\frac{{\left((4,6,0)kg.m/s\right)}^2}{2脳11kg} \\ =\frac{\left((16,36,0)kg.m/s\right)}{22kg} \\ =(0.727,1.636,0)m/s \end{gathered}$$

Thus, the initial kinetic energy of object B is $(0.727,1.636,0)m/s$.

The equation of the final kinetic energy of the object A is expressed as,

$K.E_{.A,final}=\frac{{\stackrel{鈫�}{P}}_{A,i^2}}{2m_A}$

Here, ${\stackrel{鈫�}{P}}_{A,i}$ is object A鈥檚 initial momentum is the mass of object A.

For${\stackrel{鈫�}{p}}_{A,I}=(13,3,0)kg.m/s$and $m_A=7kg$,

localid="1657863624578" $$\begin{gathered} K.E_{.A,final}=\frac{{\left((13,3,0)kg.m/s\right)}^2}{2脳7kg} \\ =\frac{\left((169,9,0)kg.m/s\right)}{14kg} \\ =(12.07,0.64,0)m/s \end{gathered}$$

Thus, the final kinetic energy of object A is $(12.07,0.64,0)m/s$.

The equation of the final kinetic energy of object B is expressed as,

$K.E_{.B,final}=\frac{{\stackrel{鈫�}{P}}_{B,i^2}}{2m_B}$

Here, ${\stackrel{鈫�}{P}}_{B,i}$ is object B鈥檚 initial momentum is the mass of object B.

For${\stackrel{鈫�}{P}}_{B,i}=(8,-2,0)kg.m/s$and$m_B=11kg$,

localid="1657864819624" $$\begin{gathered} K.E_{B,final}=\frac{{\left((8,-2,0)kg.m/s\right)}^2}{2脳11kg} \\ =\frac{{\left((64,4,0)kg.m/s\right)}^2}{22kg} \\ =(2.909,0.181,0)m/s \end{gathered}$$

Thus, the final kinetic energy of object B is $(2.909,0.181,0)m/s$.

The equation of the total initial kinetic energy of the system is expressed as,

$K.E_{.initial}=K.E_{.A,initial}+K.E_{.B,initial}$

For$K.E_{.A,initial}=(20.64,1.78,0)m/s$and$K.E_{.B,initial}=(0.727,1.636,0)m/s.$,

$$\begin{gathered} K.E_{initial}=(20.64,1.78,0)m/s+(0.727,1.636,0)m/s \\ =(21.367,3.416,0)m/s \end{gathered}$$

Thus, the total initial kinetic energy of the system is $(21.367,3.416,0)m/s$.

The equation of the total final kinetic energy of the system is expressed as,

$K.E_{.final}=K.E_{.A,final}+K.E_{.B,final}$

Forrole="math" localid="1657865760801" $K.E_{.A,final}=(12.07,0.64,0)m/s$and$K.E_{.B,final}=(12.07,0.64,0)m/s$,

$$\begin{gathered} K.E_{.A,final}=(12.07,0.64,0)m/s+(2.909,0.181,0)m/s \\ =(14.979,0.821,0)m/s \end{gathered}$$

Thus, the total final kinetic energy of the system is $(14.979,0.821,0)m/s$.

From the first law of thermodynamics, the equation of the change in the internal energy can be expressed as,

$鈭�l=鈭�Q-鈭�W$

Here, $鈭�l$is the change in the system鈥檚 internal energy, $鈭�Q$is the change in the energy flow, and $鈭�W$is the work done by the objects.

For the object A,

Rewrite the above equation by considering the system as an isolated system; that is, the value of the energy flow is 0 $鈭�Q=0$ and $鈭�W=(鈭�k.E_{.A,final}-鈭�k.E_{.B,initial})$.

role="math" localid="1657866325087" $鈭�l=0-(鈭�k.E_{.A,final}-鈭�k.E_{.A,initial})$

For$鈭�K.E_{.A,final}=(12.07,0.64,0)m/s$and$鈭�k.E_{.A,initial}=(20.64,1.78,0)m/s$,

$$\begin{gathered} 鈭�l=0-\left((12.07,0.64,0m/s-(20.64,1.78,0)m/s\right) \\ =(8.57,1.14,0)m/s \end{gathered}$$

For the object B,

The expression for the change in internal energy is as follows,

$鈭�l=0-(鈭�K.E_{B,final}-鈭�K.E_{B,initial})$

For$鈭�K.E_{B,final}=(2.909,0.181,0)m/s$and$鈭�K.E_{B,initial}=(0.727,1.636,0)m/s$,

$$\begin{gathered} 鈭�l=0-\left((2.909,0.181,0)m/s-(0.727,1.636,0)m/s\right) \\ =(-2.182,1.455,0)m/s \end{gathered}$$

Thus, the increase of internal energy of object A is $(8.57,1.14,0)m/s$and of object B is $(-2.182,1.455,0)m/s$.

The system is assumed to be an isolated system. So, the value of energy flow from the surroundings to the system is zero.

Thus, Q is assumed to be 0.