Question 5 (paper Question V) — Oblique Central Impact of Two Disks (Part B · Dynamics, 20 marks): 04-BS-3, May 2015, Q5

Question 5 (paper Question V) — Oblique Central Impact of Two Disks (Part B · Dynamics, 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. Disk A: mass 6 kg, radius 100 mm, velocity 20 m/s at 45° to the line of centres (approaching B). Disk B: mass 4 kg, radius 75 mm, velocity 10 m/s at 30° to the line of centres (approaching A). Coefficient of restitution e = 0.6. Smooth (frictionless) horizontal plane.

Given data
Quantity	Value
Mass of A	6 kg
Mass of B	4 kg
v_A before impact	20 m/s at 45°
v_B before impact	10 m/s at 30°
Coefficient of restitution e	0.6

Find. a) The velocities of A and B just after impact; b) their magnitudes and directions.

Figure 5 — oblique central impact: A (100 mm) and B (75 mm) approaching along the horizontal line of centres.

Approach. Resolve each velocity into a component along the line of centres (x, the line joining the disk centres, here horizontal) and a component tangent to it (y). The tangential components are unchanged by a smooth impact (no friction); the normal (x) components are found from conservation of momentum plus the coefficient-of-restitution equation, both applied along x only.

Resolve into components. Line of centres (x) is horizontal: $$v_{Ax}=20\cos45^\circ=14.142,\ v_{Ay}=20\sin45^\circ=14.142\ \text{m/s}$$ $$v_{Bx}=-10\cos30^\circ=-8.660,\ v_{By}=-10\sin30^\circ=-5.000\ \text{m/s}$$
Tangential components unchanged (smooth surfaces): $$v'_{Ay}=14.142\ \text{m/s}, \qquad v'_{By}=-5.000\ \text{m/s}$$
Momentum + restitution along the line of centres. $$m_Av_{Ax}+m_Bv_{Bx} = m_Av'_{Ax}+m_Bv'_{Bx}: \quad 6(14.142)+4(-8.660)=6v'_{Ax}+4v'_{Bx}$$ $$e=\dfrac{v'_{Bx}-v'_{Ax}}{v_{Ax}-v_{Bx}}=0.6=\dfrac{v'_{Bx}-v'_{Ax}}{14.142-(-8.660)}$$ Solving the 2×2 system: $v'_{Ax}=\boxed{-0.451\ \text{m/s}}$, $v'_{Bx}=\boxed{13.230\ \text{m/s}}$.
Recombine into final velocities. $$\vec{v}'_A=(-0.451,\,14.142)\ \text{m/s} \Rightarrow |\vec{v}'_A|=\boxed{14.15\ \text{m/s}}\ \text{at }91.8^\circ\text{ to the line of centres}$$ $$\vec{v}'_B=(13.230,\,-5.000)\ \text{m/s} \Rightarrow |\vec{v}'_B|=\boxed{14.14\ \text{m/s}}\ \text{at }-20.7^\circ\text{ to the line of centres}$$

Results
Quantity	Value
Velocity of A after impact	14.15 m/s, 91.8° from the line of centres (essentially straight up, tilted slightly back)
Velocity of B after impact	14.14 m/s, 20.7° below the line of centres

Topic: Oblique central impact of two smooth bodies.

Key relations: along the line of centres, $m_Av_{Ax}+m_Bv_{Bx}=m_Av'_{Ax}+m_Bv'_{Bx}$ and $e=(v'_{Bx}-v'_{Ax})/(v_{Ax}-v_{Bx})$; tangentially, $v'_{Ay}=v_{Ay}$, $v'_{By}=v_{By}$.

Why this works: a smooth (frictionless) impact can only exert force along the common normal at the contact point (the line of centres), so momentum and restitution apply only to that component; with no tangential force, each disk's tangential momentum — and hence velocity — cannot change.

Common pitfall: resolving velocities relative to the wrong axis (using the direction each disk happens to travel instead of the true line of centres), or applying the restitution equation to the resultant speeds instead of only the normal components.

Source: Hibbeler, Engineering Mechanics: Dynamics, 14th ed., Ch. 15 (Kinetics of a Particle: Impulse and Momentum).

Question 5 of 6: Question 5 (paper Question V) — Oblique Central Impact of Two Disks (Part B · Dynamics, 20 marks)

Question 5 (paper Question V) — Oblique Central Impact of Two Disks (Part B · Dynamics, 20 marks)