Question 4 of 6: Pendulum Impact on an Inclined Surface — Impulse-Momentum and Coefficient of Restitution

Question 4: Pendulum Impact on an Inclined Surface — Impulse-Momentum and Coefficient of Restitution (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Mass $m=2$ kg on a string of length $L=1$ m, pivoted at fixed point $B$; released from rest at $\theta=0^\circ$ (string horizontal, level with $B$ — the schematic angle drawn in the figure is illustrative only, the stated release angle is exactly $0^\circ$), swinging down through a quarter turn to strike a rigid, stationary incline the instant the string becomes vertical. Incline angle $\phi=60^\circ$ from horizontal; impact coefficient of restitution $e=0.7$; impact duration $\Delta t=0.05$ s; string mass negligible.

[Figure not reproduced: Pendulum released from $\theta=0^\circ$ (string horizontal) swings to vertical and strikes the $\phi=60^\circ$ incline; $X$ lies along the incline (up-slope), $Y$ is the outward normal, both confirmed against the source figure. See the official exam paper.]

Find. (a) the impulse-momentum equations for the impact, in the given $X$-$Y$ frame; (b) the velocity right after impact; (c) the average impact force over $\Delta t=0.05$ s; (d) whether (a) changes if the string is a rigid rod.

Approach. First get the pre-impact speed from energy conservation over the (string-tension-only, therefore frictionless-in-effect) swing from $\theta=0^\circ$ to vertical; resolve that velocity into components along the incline ($X$, tangential) and perpendicular to it ($Y$, normal); apply the impulse-momentum principle separately in each direction — no impulsive force tangentially (frictionless contact), and Newton's restitution rule normally — then convert back to get the impact force and examine the direction of the rebound velocity for part (d).

1. Part (a) — velocity just before impact, and the impulse-momentum equations. From $\theta=0^\circ$ (level with $B$, at rest) to the vertical position, the mass falls a height $L$ under gravity alone (the string does no work; it only redirects the velocity), so by energy conservation $$\tfrac12 m v_1^2 = mgL \ \Rightarrow\ v_1=\sqrt{2gL}=\sqrt{2(9.81)(1)}=4.429\ \text{m/s (horizontal, toward the incline).}$$ Resolving into the given axes $\hat X=(\cos\phi,\sin\phi)$ (up-slope) and $\hat Y=(-\sin\phi,\cos\phi)$ (outward normal), both read directly off the source figure: $$v_{1X}=v_1\cos\phi = 2.215\ \text{m/s},\qquad v_{1Y}=-v_1\sin\phi=-3.836\ \text{m/s}\ \ (\text{negative: approaching the surface}).$$ The impact is frictionless (nothing in the problem states otherwise, and the surface is a straight rigid incline), so no impulsive force acts tangentially; the entire impulsive reaction $J$ from the incline acts along $+Y$. The two impulse–momentum equations for the mass are therefore $$\boxed{X:\ \ m v_{1X} + 0 = m v_{2X}} \qquad\qquad \boxed{Y:\ \ m v_{1Y} + J = m v_{2Y}},$$ closed by Newton's restitution rule at the (fixed, immovable) incline, $v_{2Y}=-e\,v_{1Y}$.
2. Part (b) — velocity right after impact. From the $X$-equation, $v_{2X}=v_{1X}=2.215$ m/s (tangential component unchanged). From restitution, $v_{2Y}=-e\,v_{1Y}=-0.7(-3.836)=2.685$ m/s. So $$\boxed{\vec v_2 = 2.215\,\hat X + 2.685\,\hat Y\ \text{m/s}}, \qquad |\vec v_2|=\sqrt{2.215^2+2.685^2}=3.481\ \text{m/s},\ \ 50.5^\circ\ \text{from }\hat X\ \text{(above the surface)}.$$
3. Part (c) — average impact force. Only the $Y$-direction carries an impulsive force (the $X$-momentum is unchanged, so its average force is zero): $$J = m(v_{2Y}-v_{1Y}) = 2\big(2.685-(-3.836)\big) = 13.04\ \text{N}\cdot\text{s},\qquad \boxed{F_{\text{avg}}=\frac{J}{\Delta t}=\frac{13.04}{0.05}=260.8\ \text{N},\ \text{along }+Y\ (\text{away from the incline}).}$$ As is standard for impact problems, the weight is treated as non-impulsive: over $\Delta t=0.05$ s its impulse is only $mg\,\Delta t=2(9.81)(0.05)=0.98$ N·s, about 7.5% of $J$, and it is neglected in both equations of part (a).
4. Part (d) — would a rigid rod change (a)? Convert $\vec v_2$ back to the horizontal-vertical frame at $B$: $\vec v_2=(-1.218,\,3.261)$ m/s, i.e. it has an upward (toward-$B$) vertical component of 3.261 m/s. At the instant of impact the string is vertical, so "toward $B$" is exactly the radial direction of the pendulum's circular constraint. A flexible string can only pull (it prevents the mass from moving farther than $L$ from $B$, but exerts nothing if the mass tries to move closer): since the rebound velocity's radial component is inward (toward $B$, i.e. the "string" would need to shorten), the string simply goes slack the instant after impact and supplies zero force — it plays no role in the impact itself, and the two equations in part (a), which involve only the ball and the incline's impulsive contact force, are unaffected by whether the support is present at all. A rigid rod, by contrast, is a two-force member that can push as well as pull; it enforces the distance-from-$B$ constraint in both directions, so it would have to supply an additional impulsive (compressive) force at $B$, transmitted along the rod, to instantaneously cancel that same inward-radial velocity component and keep the mass exactly on the circle. So yes, the answer to (a) would change: with a rigid rod, a second impulsive-force term (the rod's own reaction at $B$) must be added to the momentum equations; with the actual string, no such term exists because the string cannot push, and part (a)'s two equations describe the incline impact alone.