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Define variable system; exploding projectile

Short answer

A variable system of particles gains or loses mass continuously. Momentum conservation for the explosion: 20(30) = 5 v_A cos45° + 15 v_B cos30°, 0 = 5 v_A sin45° – 15 v_B sin30°. Results: v_A = 62.12 m/s at 45° above horizontal, v_B = 29.28 m/s at 30° below.

Define variable system of particles. A 20 kg projectile is moving with a velocity of 30 m/s when it explodes into two fragments A and B, weighing 5 kg and 15 kg, respectively. Knowing that immediately after the explosion, fragments A and B travel in directions defined respectively by θA=45∘\theta_A = 45^\circ and θB=30∘\theta_B = 30^\circ, determine the velocity of each fragment. [Figure: 20 lb projectile moving right at v0=100v_0 = 100 ft/s; after explosion, 5 lb fragment A moves at θA=45∘\theta_A = 45^\circ above the horizontal and 15 lb fragment B at θB=30∘\theta_B = 30^\circ below the horizontal.]

Variable system of particles

A variable system of particles is one whose total mass changes with time because particles continuously join it or leave it, such as a rocket ejecting gas, a conveyor belt loaded with sand, or a chain being lifted from a heap. Its equation of motion is ∑F⃗=ma⃗−dmdtu⃗rel\sum\vec F=m\vec a-\dfrac{dm}{dt}\vec u_{rel}.

Exploding projectile

Internal forces of the explosion do not change the total momentum. Masses: m=20m=20 kg, mA=5m_A=5 kg, mB=15m_B=15 kg, v0=30v_0=30 m/s (the numbers given in the question text; the figure shows the same arrangement in lb and ft/s).

              A (5 kg)
               /  45 deg
 v0 = 30 --->  *----------> x
               \  30 deg
              B (15 kg)

x-direction:

20(30)=5vAcos⁡45∘+15vBcos⁡30∘20(30)=5v_A\cos45^\circ+15v_B\cos30^\circ 600=3.5355vA+12.990vB(1)600=3.5355v_A+12.990v_B\qquad(1)

y-direction:

0=5vAsin⁡45∘−15vBsin⁡30∘ ⇒ vA=2.1213vB(2)0=5v_A\sin45^\circ-15v_B\sin30^\circ\ \Rightarrow\ v_A=2.1213v_B\qquad(2)

Substituting (2) in (1):

600=7.5vB+12.990vB=20.490vB ⇒ vB=29.28 m/s600=7.5v_B+12.990v_B=20.490v_B\ \Rightarrow\ v_B=29.28\ \text{m/s} vA=2.1213(29.28)=62.12 m/sv_A=2.1213(29.28)=62.12\ \text{m/s}

Answer: vA=62.12v_A=62.12 m/s at 45∘45^\circ above the horizontal; vB=29.28v_B=29.28 m/s at 30∘30^\circ below the horizontal.

✓ Answer checked by Claude Opus 5.5 (high effort)

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