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Bullet embedded in suspended block

Short answer

Plastic impact: momentum conservation gives v = (0.02×400) / (2.02) = 3.96 m/s. Energy conservation yields rise height h = v²/(2g) = 0.799 m; cos θ = (1.5 –0.799)/1.5 gives θ = 62.2°. Kinetic energy loss is 99.0%.

A bullet of mass 20 g moving horizontally at 400 m/s strikes and remains embedded in a wooden block of mass 2 kg suspended by a string 1.5 m long. Find (a) the velocity of the block with the bullet just after impact, (b) the height to which the block rises, (c) the angle through which the string swings, and (d) the percentage of kinetic energy lost.

Given: bullet m=0.020m = 0.020 kg, u=400u = 400 m/s; block M=2M = 2 kg; string length l=1.5l = 1.5 m.

         O
         |  l
   --->  |
  bullet [M]   -> swings up through h

(a) Velocity just after impact

The impact is very short, so the string tension and weight are non-impulsive. Linear momentum is conserved horizontally:

0.020×400=(0.020+2) v  ⇒  v=8.02.02=3.960 m/s0.020 \times 400 = (0.020 + 2)\,v \;\Rightarrow\; v = \frac{8.0}{2.02} = 3.960\ \text{m/s}

(b) Height of rise

After impact, mechanical energy is conserved:

12v2=gh  ⇒  h=(3.960)22×9.81=0.799 m\tfrac12 v^2 = g h \;\Rightarrow\; h = \frac{(3.960)^2}{2 \times 9.81} = 0.799\ \text{m}

(c) Angle of swing

cos⁡θ=l−hl=1.5−0.7991.5=0.4670  ⇒  θ=62.2∘\cos\theta = \frac{l - h}{l} = \frac{1.5 - 0.799}{1.5} = 0.4670 \;\Rightarrow\; \theta = 62.2^\circ

(d) Kinetic energy lost

  • Before: 12(0.020)(400)2=1600\tfrac12(0.020)(400)^2 = 1600 J
  • After: 12(2.02)(3.960)2=15.84\tfrac12(2.02)(3.960)^2 = 15.84 J
  • Loss =1584.16= 1584.16 J, which is 1584.161600×100=99.0 %\dfrac{1584.16}{1600} \times 100 = 99.0\ \%

The loss is due to the bullet penetrating and heating the wood (plastic impact, e=0e = 0).

Answer: v = 3.96 m/s; h = 799 mm; θ = 62.2°; KE lost = 99.0 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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