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

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 ≡ U_f (X_0) − U_i (X_0).

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

\dot{P} =0.  (isolated system)

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

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

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

According to the first law (1.9)

\dot{E} =0.  (isolated system)

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

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

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.

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

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