Question 6 (paper Question VI) — Oblique Impact on an Incline (Part B · Dynamics, 20 marks): 04-BS-3, December 2015, Q6

Question 6 (paper Question VI) — Oblique Impact on an Incline (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. Free-fall height h = 0.75 m; incline angle θ = 20°; coefficient of restitution e = 0.85; impact assumed smooth (tangential velocity unchanged).

Given data
Quantity	Value
h (drop height, A to B)	0.75 m
θ (incline angle)	20°
e (coefficient of restitution)	0.85

Find. The distance L_BC along the incline from the impact point B to the landing point C.

Figure 6 — vertical drop to B, oblique bounce, and the projectile arc landing at C.

Approach. Find the impact speed from free fall, split it into components normal and tangential to the incline, apply the restitution law to the normal component only (tangential unchanged, smooth impact), convert the rebound velocity back to horizontal/vertical axes, then solve projectile motion for the point where the ball returns to the (sloped) surface.

Impact speed at B. $$v_B=\sqrt{2gh}=\sqrt{2(9.81)(0.75)}=\boxed{3.836\ \text{m/s} \downarrow}$$
Resolve into normal (n) and tangential (t) components (n perpendicular to the incline, t along the incline, positive down-slope). Since v_B is purely vertical: $$v_n=v_B\cos\theta=3.836\cos20^\circ=3.605\ \text{m/s}, \qquad v_t=v_B\sin\theta=3.836\sin20^\circ=1.312\ \text{m/s}$$
Apply the restitution law (normal) and no-friction assumption (tangential). $$v_n^\prime=e\,v_n=0.85(3.605)=\boxed{3.064\ \text{m/s (away from surface)}}, \qquad v_t^\prime=v_t=\boxed{1.312\ \text{m/s (down-slope, unchanged)}}$$
Rebound velocity in horizontal/vertical axes. With $\hat{\mathbf n}=(\sin\theta,\cos\theta)$, $\hat{\mathbf t}=(\cos\theta,-\sin\theta)$: $$\mathbf v^\prime=v_n^\prime\hat{\mathbf n}+v_t^\prime\hat{\mathbf t}=\boxed{(2.281,\ 2.430)\ \text{m/s}}$$ (v_x = 2.281 m/s, v_y = 2.430 m/s, taking B as the origin).
Projectile motion to the landing point C on the incline ($y=-x\tan\theta$ for points on the slope below B): $$x(t)=v_xt,\qquad y(t)=v_yt-\tfrac12gt^2$$ Substituting the incline condition and solving for the non-zero root: $$t=\dfrac{2(v_y+v_x\tan\theta)}{g}=\dfrac{2(2.430+2.281\tan20^\circ)}{9.81}=0.6648\ \text{s}$$ $$x_C=v_xt=2.281(0.6648)=1.516\ \text{m}, \qquad y_C=-x_C\tan\theta=-0.552\ \text{m}$$
Distance along the incline. $$L_{BC}=\dfrac{x_C}{\cos\theta}=\dfrac{1.516}{\cos20^\circ}=\boxed{1.613\ \text{m}}$$

Final results
Quantity	Value
Impact speed v_B	3.836 m/s
Rebound velocity (v_x, v_y)	(2.281, 2.430) m/s
Time of flight B to C	0.665 s
L_BC	1.613 m

Topic: Oblique (non-central) impact of a particle with a fixed surface, followed by projectile motion.

Key relations: restitution $e=-v_n^\prime/v_n$ along the line of impact (the surface normal, for impact with a fixed massive surface); tangential momentum/velocity is preserved when the surface is smooth; standard projectile kinematics $x=v_xt$, $y=v_yt-\tfrac12gt^2$.

Why this works: the impulsive contact force during a short impact acts along the common normal, so only the normal velocity component is affected by the collision law; decomposing into normal/tangential axes isolates that effect cleanly, and re-projecting onto horizontal/vertical axes hands the problem back to ordinary projectile motion under gravity alone.

Common pitfall: applying the restitution factor e to the whole impact speed v_B instead of only its normal component, or mixing up which axis is n and which is t when resolving a purely vertical incoming velocity.

Source: Hibbeler, Engineering Mechanics: Dynamics, 14th ed., Ch. 15 (Kinetics of a Particle: Impulse and Momentum § Oblique Impact) and Ch. 12 (Kinematics of a Particle § Projectile Motion).

Question 6 of 6: Question 6 (paper Question VI) — Oblique Impact on an Incline (Part B · Dynamics, 20 marks)

Question 6 (paper Question VI) — Oblique Impact on an Incline (Part B · Dynamics, 20 marks)