AP Physics 1 4.4: Elastic and Inelastic Collisions

(iii) For a multistep derivation that includes the correct relationship between kinetic energy,

mass, and speed (i.e., $K=\frac{1}{2} m v^{2}$ )

Scoring Note: The minimum requirement to earn this point is to show an expression of

kinetic energy or change in energy that goes beyond the given equation on the reference

information. For example, substituting quantities from the problem into $K=\frac{1}{2} m v^{2}$ or

indicating the change in kinetic energy is $\Delta K=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}$ earns the point.

For an indication that $m_{\mathrm{c}}$ and $\frac{6}{5} m_{\mathrm{c}}$ are the initial and final masses, respectively, of the

objects moving horizontally

For including an expression for $v_{f}$ consistent with part A (ii)

Scoring Note: A correct, isolated, final expression of $\Delta K=-\frac{1}{12} m_{\mathrm{c}} v_{\mathrm{c}}^{2}$ earns points

A3, A4, A6, and A7.

Example Response

$$\begin{aligned} \Delta K & =K_{f}-K_{0} \\ K_{0} & =\frac{1}{2} m_{\mathrm{c}} v_{\mathrm{c}}^{2} \\ K_{f} & =\frac{1}{2}\left(\frac{6}{5} m_{\mathrm{c}}\right)\left(\frac{5}{6} v_{\mathrm{c}}\right)^{2}=\frac{5}{12} m_{\mathrm{c}} v_{\mathrm{c}}^{2} \\ \Delta K & =-\frac{1}{12} m_{\mathrm{c}} v_{\mathrm{c}}^{2} \end{aligned}$$