# Change in temperature after an inelastic collision

• powerof
In summary, at the highest point the object (bullet+sphere) is at, the projectile has a kinetic energy of 2gh and the sphere has a kinetic energy of 1gh. The temperature of the object will increase by 1degree Celsius.

## 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.

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

Have a nice day.

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:

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

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