---
title: "Bullet embedded in suspended block"
subject: "Engineering Mechanics"
chapter: 4
key: "semester_3/engineering_mechanics:ballistic-pendulum-bullet-block"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/ballistic-pendulum-bullet-block
site: BCT Engineering
---

# Bullet embedded in suspended block

Engineering Mechanics · Chapter 4: Kinetics of Particles: Impulse and Momentum

- Asked in: year not given, 6 marks
- Video: [Ballistic Pendulum Physics Problems - Conservation of Momentum & Energy - Inelastic Collisions](https://www.youtube.com/watch?v=PUNKqLkELbQ) — The Organic Chemistry Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/ballistic-pendulum-bullet-block
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-4

## Question

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.

## 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%.

## Answer

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

```text
         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:

$$
0.020 \times 400 = (0.020 + 2)\,v \;\Rightarrow\; v = \frac{8.0}{2.02} = 3.960\ \text{m/s}
$$

#### (b) Height of rise
After impact, mechanical energy is conserved:

$$
\tfrac12 v^2 = g h \;\Rightarrow\; h = \frac{(3.960)^2}{2 \times 9.81} = 0.799\ \text{m}
$$

#### (c) Angle of swing

$$
\cos\theta = \frac{l - h}{l} = \frac{1.5 - 0.799}{1.5} = 0.4670
\;\Rightarrow\; \theta = 62.2^\circ
$$

#### (d) Kinetic energy lost
- Before: $\tfrac12(0.020)(400)^2 = 1600$ J
- After: $\tfrac12(2.02)(3.960)^2 = 15.84$ J
- Loss $= 1584.16$ J, which is $\dfrac{1584.16}{1600} \times 100 = 99.0\ \%$

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

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which quantity is conserved during the bullet's embedding?
   A. Mechanical energy
   B. Horizontal linear momentum
   C. Kinetic energy
   D. Angular velocity
   Answer: B. The impact is plastic so KE is lost, but momentum is conserved because string tension and weight are non-impulsive. (Key point: What is conserved in impact)

2. The block velocity just after impact is:
   A. 400 m/s
   B. 0.02 x 400 / 2.02 = 3.96 m/s
   C. 400 / 2 = 200 m/s
   D. 0.02 x 400 / 2 = 4 m/s
   Answer: B. Bullet and block move together, so total mass 2.02 kg is used. (Key point: Velocity after impact)

3. After impact, how are the rise and angle found?
   A. Energy conservation gives h = v^2/2g = 0.799 m; cos(theta) = (l - h)/l gives 62.2 deg
   B. Momentum gives h; theta = h/l
   C. h = v^2/g; theta = sin^-1(h/l)
   D. Energy gives h = 2v^2/g
   Answer: A. Swing is a conservative motion, then geometry of the string gives the angle. (Key point: Height and angle)

4. About what percentage of the bullet's kinetic energy is lost?
   A. 1 %
   B. 50 %
   C. 99 %
   D. 0 %
   Answer: C. 1584.16 J of 1600 J is lost (99.0 %) because the impact is plastic. (Key point: Energy lost)

## Related in Engineering Mechanics

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## Related in other subjects

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