Question 4 of 6: Bullet-Block Impact on a Smooth Incline

Question 4: Bullet-Block Impact on a Smooth Incline (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. $m_A=2\text{ lb}$, $m_B=0.1\text{ lb}$, incline angle $\theta=30^\circ$, $\vec v_B^-=200\text{ m/s}$ horizontal, $\Delta t=0.015\text{ s}$, block starts at rest and is kinematically confined to the incline (so its velocity has no component perpendicular to the incline, at any time).

Find. (a),(b) the symbolic impulse–momentum vector equations; (c),(d) which forces are impulsive and why weight is negligible; (e) $\vec v_A^+$ and $F_{imp}$ (numeric); (f) the restitution equation (symbolic).

Approach. Resolve the bullet's horizontal velocity into the incline's local $x$–$y$ axes; apply impulse–momentum first to the COMBINED bullet+block system (the unknown internal impact force cancels, isolating the incline's normal reaction as the only external impulsive force, purely along $y$) to get $v_A^+$; then apply it to the bullet ALONE (whose only impulsive force is the reaction from the block) to get $F_{imp}$.

Part (a) — block $A$ alone. With $\vec v_A^-=\vec 0$ and forces $\vec F_{imp}$ (impact reaction from the bullet), $N\vec j$ (incline normal reaction) and $R_S\vec i$ (stop-block reaction): $$\boxed{m_A\vec v_A^- + \int\vec F_{imp}\,dt + \int N\,dt\,\vec j + \int R_S\,dt\,\vec i = m_A v_A^+\vec i}$$

Part (b) — bullet $B$ and block $A$ together. The internal impact-force pair cancels by Newton's third law, leaving only the incline's normal reaction as an external impulsive force (the bullet itself never touches the incline): $$\boxed{m_B\vec v_B^- + m_A\vec v_A^- + \int N\,dt\,\vec j = (m_A+m_B)\,v_A^+\vec i}$$ (both bodies share the single post-impact velocity $v_A^+\vec i$ since the bullet embeds and the combined mass is confined to the incline).

Part (c) — which forces turn impulsive? The bullet's very large, very brief contact force $F_{imp}$ is impulsive by definition. Numerically (Part e) the impact drives the block UP-slope, away from the stop $S$ — and a stop can only push, never pull — so the stop reaction $R_S$ drops to (and stays at) zero: it does not become impulsive. The incline's normal force $N$, by contrast, does become impulsive: the block is kinematically forced to keep zero velocity perpendicular to the incline throughout, so $N$ must instantaneously cancel the bullet's large impulsive $y$-momentum contribution — an ordinary, finite contact force could never do that in zero time.

Part (d) — why ignore the weights. A weight's impulse over the impact is $\int m g\,dt = mg\,\Delta t$, an ordinary, FINITE force multiplied by a vanishingly short time: its impulse $\to0$ as $\Delta t\to0$. $F_{imp}$'s impulse stays finite only because $F_{imp}$ itself is enormous (it is what makes it "impulsive" in the technical sense) — the weights never grow to compensate, so their contribution is negligible next to $F_{imp}\Delta t$.

1. Resolve the bullet's velocity into the incline's local axes. With $\hat x=(\cos\theta,\sin\theta)$ (up-slope) and $\hat y=(-\sin\theta,\cos\theta)$ (outward normal), and $\vec v_B^-=(200,0)$ globally: $$V_{Bx}^- = 200\cos30^\circ = 173.21\text{ m/s},\qquad V_{By}^- = -200\sin30^\circ = -100.0\text{ m/s}$$
2. Solve Part (b) for $v_A^+$ along $x$ (mass RATIO is unit-independent, so the mixed lb/SI units cause no error here): $$v_A^+ = \frac{m_B}{m_A+m_B}\,V_{Bx}^- = \frac{0.1}{2.1}(173.21) = \boxed{8.25\text{ m/s (up-slope)}}$$
3. Solve the bullet-alone equation for $F_{imp}$ (here the mass VALUES matter, so $m_B=0.1\text{ lb}=0.04536\text{ kg}$): $$\vec F_{imp} = \frac{m_B\big[(V_{Bx}^--v_A^+)\hat x + V_{By}^-\hat y\big]}{\Delta t}$$ $$F_{imp,x} = \frac{0.04536(173.21-8.25)}{0.015} = 498.8\text{ N},\qquad F_{imp,y} = \frac{0.04536(-100.0)}{0.015} = -302.4\text{ N}$$ $$\boxed{F_{imp} = \sqrt{498.8^2+302.4^2} = 583.3\text{ N}}$$

Part (f) — restitution if the bullet bounces. The physical line of impact is the bullet's original horizontal line of flight, so restitution must compare velocity components along the GLOBAL horizontal, not the local $x$-axis. Projecting the local-frame components onto that direction ($V\cos\theta - V_y\sin\theta$ for a local vector $(V_x,V_y)$, since global-$X=\hat x\cos\theta - \text{...}$, worked out from $\hat x,\hat y$ above) and forming $e=-\dfrac{(v_2'-v_1')_n}{(v_2-v_1)_n}$ with $v_{A}^-=0$: $$\boxed{e = \dfrac{V_{Ax}^+\cos\theta - V_{Bx}^+\cos\theta + V_{By}^+\sin\theta}{V_{Bx}^-\cos\theta - V_{By}^-\sin\theta}}$$