Oblique Impact of a Falling Ball on an Incline (Part B — Dynamics): 04-BS-3, May 2013, Q5

Question V: Oblique Impact of a Falling Ball 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. Ball mass 2 kg, released from rest, falls $h_1=3$ m to reach the 45° smooth incline at B. Coefficient of restitution $e=0.8$. Vertical drop from B to the base plane is $h_2=2$ m.

Given data
Quantity	Value
Free-fall height to B	3 m
Incline angle	45° (smooth)
Restitution $e$	0.8
Drop from B to base	2 m

Figure 5 — ball free-falls 3 m to B, rebounds off the smooth 45° incline, and lands at A on the horizontal plane.

Find. (a) The horizontal distance $s$ (measured from the base of the incline) where the ball lands at A. (b) The ball's velocity at A.

Approach. Find the impact speed at B from free fall, resolve it into components normal and tangential to the smooth incline, apply $e$ to the normal component only (tangential is unchanged — the surface is smooth), then treat the rebound as ordinary projectile motion from B to the horizontal plane at A.

Speed at impact (B). $v_0=\sqrt{2g h_1}=\sqrt{2(9.81)(3)}=7.672$ m/s, straight down.
Resolve into normal/tangential components. With the incline tangent $\hat t=(\cos(-45^\circ),\sin(-45^\circ))$ and normal $\hat n=(\cos45^\circ,\sin45^\circ)$, $$v_{0n}=\mathbf v_0\cdot\hat n=-5.425\text{ m/s}, \qquad v_{0t}=\mathbf v_0\cdot\hat t=5.425\text{ m/s}$$
Apply restitution (normal) and smoothness (tangential unchanged). $v_{1n}=-e\,v_{0n}=0.8(5.425)=4.340$ m/s (away from the surface), $v_{1t}=v_{0t}=5.425$ m/s. Recombining, $$\mathbf v_1 = v_{1n}\hat n+v_{1t}\hat t = \boxed{(6.905,\,-0.767)\text{ m/s}}$$
Projectile motion from B to A. The ball must fall a further $h_2=2$ m: solving $\tfrac12 g t^2 - v_{1y}t - h_2=0$ (with $v_{1y}=-0.767$ m/s) for the positive root gives $t=0.565$ s. Horizontal travel from B is $\Delta x=v_{1x}t=6.905(0.565)=3.902$ m. Since the incline itself runs 2 m horizontally under its 45° slope (equal rise and run), $$s=\Delta x - h_2 = 3.902-2.000=\boxed{1.90\text{ m}}$$
Velocity at A. $v_{Ax}=v_{1x}=6.905$ m/s (unchanged), $v_{Ay}=v_{1y}-gt=-0.767-9.81(0.565)=-6.311$ m/s. $$|\mathbf v_A|=\sqrt{6.905^2+6.311^2}=\boxed{9.35\text{ m/s}}, \quad 42.4^\circ\text{ below horizontal}$$

Quantity	Value
Speed at B (impact)	7.67 m/s
Velocity after rebound	(6.91, −0.77) m/s
Distance $s$	1.90 m
Speed at A	9.35 m/s (42.4° below horizontal)

Concept & Theory

Topic. Oblique (non-central) impact on a fixed smooth surface, combined with projectile motion.

Key relations. Restitution along the line of impact (the surface normal): $e=-v_{1n}/v_{0n}$. Smooth surface: tangential momentum/velocity unchanged. Projectile motion: $x=v_xt$, $y=v_yt-\tfrac12gt^2$.

Why this works. A smooth surface can exert no force along its own face, so only the velocity component perpendicular to it can change during impact; that component reverses and shrinks by the restitution factor $e$, while the parallel component simply carries through, after which ordinary constant-acceleration kinematics governs the flight to A.

Common pitfall. Applying $e$ to the whole speed instead of just the normal component; and forgetting that the horizontal distance $s$ is measured from the toe of the incline, not from the impact point B's own horizontal position, so the incline's own 2 m horizontal run must be subtracted.

Source. Hibbeler, Engineering Mechanics: Dynamics (14th ed.), Ch. 15 — Kinetics of a Particle: Impulse and Momentum, §15.4 Impact.

Question 5 of 6: Oblique Impact of a Falling Ball on an Incline (Part B — Dynamics)

Question V: Oblique Impact of a Falling Ball on an Incline (Part B — Dynamics) (20 marks)