Skip to main content

Projectile exploding into two fragments (20 lb)

Short answer

Using conservation of linear momentum in x and y: 20(100) = 5 v_A cos45° + 15 v_B cos30°, 0 = 5 v_A sin45° – 15 v_B sin30°. Solving yields v_A = 207.1 ft/s at 45° above the horizontal, v_B = 97.6 ft/s at 30° below the horizontal.

A 20-lb projectile is moving with a velocity of 100 ft/s when it explodes into two fragments A and B, weighing 5 lb and 15 lb, 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.]

The explosion forces are internal, so the total linear momentum is conserved. Weights are proportional to masses, so the factor 1/g1/g cancels and the weights can be used directly.

              A (5 lb)
               /  45 deg
 v0 =100 --->  *----------> x
               \  30 deg
              B (15 lb)

Before: the projectile moves horizontally at 100100 ft/s.

x-direction:

20(100)=5vAcos⁡45∘+15vBcos⁡30∘20(100)=5v_A\cos45^\circ+15v_B\cos30^\circ 2000=3.5355vA+12.990vB(1)2000=3.5355v_A+12.990v_B\qquad(1)

y-direction (initial vertical momentum is zero):

0=5vAsin⁡45∘−15vBsin⁡30∘0=5v_A\sin45^\circ-15v_B\sin30^\circ 3.5355vA=7.5vB ⇒ vA=2.1213vB(2)3.5355v_A=7.5v_B\ \Rightarrow\ v_A=2.1213v_B\qquad(2)

Substituting (2) in (1):

2000=3.5355(2.1213vB)+12.990vB=20.490vB2000=3.5355(2.1213v_B)+12.990v_B=20.490v_B vB=97.61 ft/s,vA=2.1213(97.61)=207.06 ft/sv_B=97.61\ \text{ft/s},\qquad v_A=2.1213(97.61)=207.06\ \text{ft/s}

Answer: vA=207.1v_A=207.1 ft/s at 45∘45^\circ above the horizontal; vB=97.6v_B=97.6 ft/s at 30∘30^\circ below the horizontal.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Applied Mechanics (Dynamics) chapter 3 questions →