Question 5 (paper Question V) — Pendulum Impact and Projectile Trajectory (Part B · Dynamics, equal value): 04-BS-3, May 2017, Q5

Question 5 (paper Question V) — Pendulum Impact and Projectile Trajectory (Part B · Dynamics, equal value)

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. Sphere A ($m=0.150$ kg) is released from rest at $\theta_1=20^\circ$ from vertical on a string of length $l=900$ mm. Sphere B ($m=0.150$ kg) sits at rest at the table edge, on the pendulum's arc at $\theta_2=18^\circ$ from vertical — on the opposite side of vertical from the release point (confirmed by the figure's rising-then-falling trajectory for B, which requires an upward velocity component at impact). Coefficient of restitution $e=0.92$; friction and string mass are neglected.

Find. The horizontal range $x$ and maximum height $y$ of sphere B's trajectory after impact.

Figure 5 (reconstructed). Sphere A released at 20°, striking sphere B at 18° on the far side of vertical; B's subsequent projectile arc.

Approach. Use energy conservation to find A's speed just before impact; model the impact as direct central (equal masses, one initially at rest) to find B's launch velocity along the line of A's motion; then apply projectile motion (launch and landing at the same height) to find B's range and peak height.

Speed of A just before impact (energy conservation). A descends from $\theta_1=20^\circ$ to $\theta_2=18^\circ$ (still swinging toward, and just past, the bottom of the arc), a net height drop of $l(\cos\theta_2-\cos\theta_1)$: $$v_A=\sqrt{2gl(\cos18^\circ-\cos20^\circ)}=\sqrt{2(9.81)(0.9)(0.0114)}=\boxed{0.448\ \text{m/s}}$$ directed tangent to the arc at $\theta_2$, i.e. up and to the left at 18° above horizontal (the swing has passed the bottom and is rising up the far side).
Direct central impact (equal masses, B initially at rest). For $m_A=m_B$: $$v_A'=\dfrac{(1-e)}{2}v_A=0.0179\ \text{m/s}, \qquad v_B'=\dfrac{(1+e)}{2}v_A=\boxed{0.430\ \text{m/s}}$$ both directed along the same line as A's incoming velocity (up-left at 18° above horizontal).
Launch velocity components of B. $v_{Bx}=-v_B'\cos18^\circ=-0.4090\ \text{m/s}$ (leaving the table edge), $v_{By}=v_B'\sin18^\circ=0.1329\ \text{m/s}$ (upward).
Projectile motion (launch and landing at the same height). $$y=\dfrac{v_{By}^2}{2g}=\dfrac{0.1329^2}{2(9.81)}=\boxed{0.900\ \text{mm}}$$ $$x=\dfrac{v_B'^2\sin(2\times18^\circ)}{g}=\dfrac{0.430^2\sin36^\circ}{9.81}=\boxed{11.08\ \text{mm}}$$

Final results — Question 5
Quantity	Value
Speed of A just before impact	0.448 m/s
Speed of B just after impact	0.430 m/s
Maximum height $y$	0.900 mm
Horizontal range $x$	11.08 mm

Topic: Energy methods, direct central impact, and projectile motion in sequence.
Key relations: $\tfrac12mv^2=mgh$ (pendulum), $e=\dfrac{v_B'-v_A'}{v_A-v_B}$ with momentum conservation for impact, $y=v_{y0}^2/2g$ and $x=v_0^2\sin2\theta/g$ for a same-height projectile.
Why this works: the three physical stages (swing, impact, flight) are independent and are chained by using each stage's output as the next stage's input; the equal-mass restitution formulas reduce a general two-body impact to two simple scalar expressions.
Common pitfall: assuming impact happens at the bottom of the swing ($\theta=0$); here the figure places B on the far side at 18°, so A is still moving (and rising) at impact, and the impact velocity has both magnitude (from the full 20°→18° height change) and a non-horizontal direction that must be carried into the projectile stage.
Source: Hibbeler, Engineering Mechanics: Dynamics (14th ed.), Ch. 14 (Work and Energy) and Ch. 15 (Impact).

Question 5 of 6: Question 5 (paper Question V) — Pendulum Impact and Projectile Trajectory (Part B · Dynamics, equal value)

Question 5 (paper Question V) — Pendulum Impact and Projectile Trajectory (Part B · Dynamics, equal value)