Kevin S. Huang

2015: Problems 9-10

Topic: Collisions
Concepts: Conservation of linear momentum, Vectors

Solution:

By conservation of linear momentum,

\vec{p}_1+\vec{p}_2=\vec{p}_i

Since $\vec{p}_1$ makes a right angle with $\vec{p}_2$ , the three momenta vectors form the sides of a right triangle. Thus,

p_1^2+p_2^2=p_i^2

(m_1v_{1f})^2+(m_2v_2)^2=(m_1v_{1i})^2

Solving for $v_{1f}$ ,

v_{1f}^2+\frac{m_2^2v_2^2}{m_1^2}=v_{1i}^2

v_{1f}=\sqrt{v_{1i}^2-\left(\frac{m_2v_2}{m_1}\right)^2}=\sqrt{(5\,\mathrm{m/s})^2-\left[\frac{0.750\,\mathrm{kg}(4\,\mathrm{m/s})}{0.650\,\mathrm{kg}}\right]^2}=1.92\,\mathrm{m/s}

so the answer is A.

Topic: Energy
Concepts: Kinetic energy

Solution:

The initial kinetic energy is given by

K_i=\frac{1}{2}m_1v^2_{1i}=\frac{1}{2}(0.650\,\mathrm{kg})(5\,\mathrm{m/s})^2=8.13\,\mathrm{J}

We found in the previous problem that $v_{1f}=1.92\,\mathrm{m/s}$ . The final kinetic energy is then given by

K_f=\frac{1}{2}m_1v^2_{1f}+\frac{1}{2}m_2v^2_2=\frac{1}{2}(0.650\,\mathrm{kg})(1.92\,\mathrm{m/s})^2+\frac{1}{2}(0.750\,\mathrm{kg})(4\,\mathrm{m/s})^2=7.20\,\mathrm{J}

and

\Delta K=K_f-K_i=-0.93\,\mathrm{J}

Thus, we have

-K_i<\Delta K<0

so the answer is E.