Impulse and Momentum Questions and Answers

Answer: C

The momentum just after impact is \(17.0\,\hat{\imath} + 9.8\,\hat{\jmath}\) kg m/s.

Speed at impact and momentum, pointing straight down:

\[ \begin{aligned} v &= \sqrt{2gh} = \sqrt{2 \times 9.8 \times 4.9} \\ &= 9.8\ \text{m/s} \\ \vec{p} &= 19.6\ \text{kg m/s, downward} \end{aligned} \]

In a perfectly elastic impact with a fixed smooth surface, the component of momentum along the surface is kept and the component along the normal is reversed. The face falls to the right at 30°, so its outward normal is \(\hat{n} = \sin 30^{\circ}\,\hat{\imath} + \cos 30^{\circ}\,\hat{\jmath}\). Then

\[ \begin{aligned} \vec{p}\cdot\hat{n} &= -19.6 \times 0.866 = -16.97 \\ \vec{p}' &= \vec{p} - 2(\vec{p}\cdot\hat{n})\,\hat{n} \\ &= -19.6\,\hat{\jmath} \\ &\quad + 33.95(0.5\,\hat{\imath} + 0.866\,\hat{\jmath}) \\ &= 16.97\,\hat{\imath} + 9.8\,\hat{\jmath} \end{aligned} \]

The size stays 19.6, as it must when no energy is lost (\(16.97^{2} + 9.8^{2} = 19.6^{2}\)). \(19.6\,\hat{\jmath}\) is the rebound from a horizontal floor, and \(19.6\,\hat{\imath}\) a deflection through 90°. \(9.8\,\hat{\imath} + 17.0\,\hat{\jmath}\) swaps the components, which is what a 60° face would give.

Answer: A

Her angular velocity increases.

The ice exerts almost no torque about the vertical axis, so the skater's angular momentum \(L = I\omega\) is conserved. Pulling her arms and legs in brings mass closer to the axis and reduces her moment of inertia I, so \(\omega\) must increase to keep \(I\omega\) constant.

A decrease would happen if she extended her limbs. The angular velocity would stay the same only if I were unchanged, and it would go to zero only under an external braking torque.