Question 6 (paper Question VI) — Projectile Fragmentation (Part B · Dynamics, 20 marks): 04-BS-3, December 2013, Q6

Question 6 (paper Question VI) — Projectile Fragmentation (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. Projectile mass 5 kg, horizontal velocity 700 m/s just before exploding; fragment A: mass 2 kg, departs at 45° above horizontal; fragment B: mass 3 kg, departs at 30° below horizontal; the ground (where both fragments land) is 60 m below the explosion point.

Given data
Quantity	Value
Projectile mass, velocity	5 kg, 700 m/s (horizontal)
Fragment A mass, angle	2 kg, 45° above horizontal
Fragment B mass, angle	3 kg, 30° below horizontal
Drop to ground	60 m

Find. Speeds v_A, v_B just after the explosion, and the horizontal distance d_A from the explosion point to where fragment A lands.

Figure 6 — explosion point and the two fragment trajectories to the (common) ground level.

Approach. The explosion is an internal event (no external horizontal or vertical impulse in that instant), so total linear momentum is conserved across the split; solving the resulting 2-equation system gives both fragment speeds. Fragment A's landing distance then follows from ordinary projectile motion using its own launch velocity and the 60 m drop.

Conservation of momentum (x: along the original 700 m/s direction; y: perpendicular, originally zero): $$m v_0 = m_A v_A\cos45^\circ + m_B v_B\cos30^\circ:\quad 5(700) = 2v_A\cos45^\circ+3v_B\cos30^\circ$$ $$0 = m_A v_A\sin45^\circ - m_B v_B\sin30^\circ:\quad 2v_A\sin45^\circ = 3v_B\sin30^\circ$$
Solve simultaneously (2 equations, 2 unknowns): $$v_A = \boxed{905.9\ \text{m/s}}, \qquad v_B = \boxed{854.1\ \text{m/s}}$$
Fragment A's velocity components and the time to fall 60 m below the explosion point: $$v_{Ax}=v_A\cos45^\circ = 640.5\ \text{m/s}, \qquad v_{Ay}=v_A\sin45^\circ = 640.5\ \text{m/s (up)}$$ $$-60 = v_{Ay}t - \tfrac12 g t^2 \Rightarrow t = \boxed{130.68\ \text{s}}$$
Horizontal distance to C: $$d_A = v_{Ax}\,t = 640.5(130.68) = \boxed{83{,}709\ \text{m}}$$ The very long flight time and distance follow directly from fragment A's huge upward speed component (640 m/s) after the explosion — it must climb, decelerate under gravity, and fall back through more than its own launch height before it can reach the (comparatively small) 60 m net drop, all while travelling horizontally the whole time.

Fragment velocities and landing distance
Quantity	Value
v_A	905.9 m/s
v_B	854.1 m/s
Time of flight for A	130.7 s
d_A	83,709 m

Topic: Conservation of linear momentum for an explosive separation, combined with projectile motion.

Key relations: $\sum m_i\mathbf{v}_i$ (before) $=\sum m_i\mathbf{v}_i$ (after); projectile kinematics $y=y_0+v_{y0}t-\tfrac12gt^2$, $x=v_{x0}t$.

Why this works: the explosive forces are internal and act over a negligible time, so they produce no net external impulse on the system — total momentum (both components) is identical immediately before and after the split, giving exactly two independent scalar equations for the two unknown fragment speeds.

Common pitfall: mixing up which fragment's angle is measured above vs. below the original flight line when writing the y-momentum equation (a sign error here silently swaps the two speeds).

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

Question 6 of 6: Question 6 (paper Question VI) — Projectile Fragmentation (Part B · Dynamics, 20 marks)

Question 6 (paper Question VI) — Projectile Fragmentation (Part B · Dynamics, 20 marks)