We consider two colliding objects that remain attached after impact. It can be shown that this type of collision has the maximum kinetic energy change. We treat the colliding objects as two point masses M1 and M2 that constitute an isolated system. The point mass M1 has an initial momentum P1 and

Question

We consider two colliding objects that remain attached after impact. It can be shown that this type of collision has the maximum kinetic energy change.We treat the colliding objects as two point masses $M_1$ and $M_2$ that constitute an isolated system. The point mass $M_1$ has an initial momentum $P_1$ and the point mass $M_2$ is initially at rest. The state variables are the momentum and the extensive variable $X_0$ associated to an internal property of the system (i.e. the entropy S as we will see in the following chapter). Let $E_i (P, X_0)$ be the energy and $U_i (X_0)$ the internal energy just before the impact. Let $E_f (P, X_0)$ be the energy and $U_f (X_0)$ be the internal energy just after the impact. Using the conservation laws of energy (1.9)

$\dot{E} =0$ . (isolated system)

and momentum (1.12),

$\dot{P} =0$ . (isolated system)

determine the variation of internal energy of the system $ΔU \equiv U_f (X_0) - U_i (X_0)$ .

[latex]\dot{E} =0[/latex]. (isolated system)

and momentum (1.12),

[latex]\dot{P} =0[/latex]. (isolated system)

determine the variation of internal energy of the system [latex]ΔU ≡ U_f (X_0) − U_i (X_0)[/latex].

Accepted Answer

According to the first law (1.12) for an isolated system

[latex]\dot{P} =0[/latex]. (isolated system)

, the momentum P is a constant, i.e. [latex]P = P_1[/latex]. The energy of the system before and after the impact reads,

[latex]E_i (P, X_0) = \frac{P^{2}_{1}}{2M_1} + U_i (X_0)[/latex].

[latex]E_f (P, X_0) = \frac{P^{2}_{1}}{2(M_1+M_2)} + U_f (X_0)[/latex]

According to the first law (1.9)

[latex]\dot{E} =0[/latex]. (isolated system)

for an isolated system, the energy [latex]E (P, X_0)[/latex] is also a constant, i.e. [latex]E_i (P, X_0) = E_f (P, X_0)[/latex]. This is expressed as,

[latex]\frac{P^{2}_{1}}{2M_1} + U_i (X_0)=\frac{P^{2}_{1}}{2(M_1+M_2)} + U_f (X_0)[/latex].

[latex]ΔU=U_f (X_0)-U_i (X_0)=\frac{P^{2}_{1} }{2} (\frac{1}{M_1}-\frac{1}{M_1+M_2} )>0[/latex].

In conclusion, the internal energy of this isolated system increases during the collision in order to compensate for the loss of kinetic energy.

Question 1.10.3: We consider two colliding objects that remain attached after...

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