Change in temperature after an inelastic collision

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SUMMARY

The discussion focuses on calculating the change in temperature after an inelastic collision between a sphere and a bullet. The sphere has a mass of 0.75 kg, while the bullet has a mass of 0.015 kg and an initial velocity of 300 m/s. The conservation of momentum and mechanical energy principles are applied to determine the height reached by the combined mass post-collision and the heat generated, which is calculated using the formula ΔT=Q/(mc) for both materials involved. The specific heat capacities are noted as 0.03 cal/(g ºC) for the sphere and 0.12 cal/(g ºC) for the bullet.

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  • Understanding of conservation of momentum in inelastic collisions
  • Knowledge of mechanical energy conservation principles
  • Familiarity with specific heat capacity and heat transfer calculations
  • Basic algebra for solving equations involving multiple variables
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Homework Statement



We have a sphere of mass M=0.75 Kg hung from the ceiling by a massless tense cord. Said sphere is hit by a projectile/bullet of mass m=0.015 Kg with velocity v_{0}=300 m/s and they are stuck together (inelastic collision). Find the height the (projectile+sphere) object reaches and whether it will get heated (ie, is there any heat produced by the impact). If so, calculate the change in the temperature of the object. The bullet and the sphere both have the same temperature at the moment of impact.

Known data: M, m, v_{0} and the specific heat capacities: c(sphere): 0.03 cal/(g ºC), c(bullet): 0.12 cal/(g ºC).

Homework Equations



Q=mcΔT, p=mv, KE=.5mv^2, PE ≈ mgh, ME=KE+PE

The Attempt at a Solution



Concerning the height, it is relatively easy to solve. The net force on the system is (assumed to be) zero, therefore the lineal momentum is conserved:

mv_{0}+M \cdot 0=(m+M)v_{f}\Rightarrow v_{f}=\frac{mv_{0}}{(m+M)}

From the moment of collision onwards the mechanical energy is conserved since only conservative forces (gravity) affect the system. Also at the highest point the kinetic energy is null (for the potential energy I take as a reference point h=0 the initial level/height of the bullet and sphere):

\frac{1}{2}(m+M)v_{f}^{2}+mg \cdot 0=\frac{1}{2}(m+M) \cdot 0+(m+M)gh \Rightarrow v_{f}^{2}=2gh \Rightarrow (\frac{mv_{0}}{(m+M)})^2=2gh\Rightarrow h= \frac{1}{2g}(\frac{mv_{0}}{(m+M)})^2

(note that v_{f} is actually the initial velocity but I named it this way to have continuity between this and the previous equation)

My problem is with the heat. Assuming that all the difference in kinetic energy goes into producing heat (instead of other things as sound, etc.) then, since the initial and final kinetic energies differ:

KE (initial) + A = KE (final) → A=ΔKE.

I think this energy A is converted or manifests itself as heat instead of kinetic energy (assuming 100% efficiency), but how do I calculate the temperature? From the formula ΔT=Q/(mc) but that is assuming a uniform material; the object is made of different materials.

Please give me some tips/pointers.

Thank you for reading, and let me know if you want me to clarify anything.

Have a nice day.
 
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Yes, so you have two materials that get heated. Two c values, two Q values, one ΔT (the bullet is inside, so the temperature difference between sphere and bullet evens out).
 
Of course, it's now so obvious. We just have a system:

\left\{\begin{matrix}\Delta T=\frac{Q_{1}}{m_{1}c_{1}}=\frac{Q_{2}}{m_{2}c_{2}}<br /> \\ Q_{1} + Q_{2} = A = \Delta (E_{K})<br /> <br /> \end{matrix}\right. \Rightarrow \frac{\Delta (E_{K})-Q_{2}}{m_{1}c_{1}}=\frac{Q_{2}}{m_{2}c_{2}} \Rightarrow \Delta (E_{K})=(\frac{m_{1}c_{1}}{m_{2}c_{2}}+1)\cdot Q_{2} \Rightarrow \frac{\Delta E_{K}}{(\frac{m_{1}c_{1}}{m_{2}c_{2}}+1)}=Q_{2}

After having gotten Q2 we just substitute in the formula for ΔT and we get the desired answer.

Thanks for your help. I realized the solution while reading your post.
 

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