---
title: "Bullet embedded in bar, h = 240 mm"
subject: "Applied Mechanics (Dynamics)"
chapter: 6
key: "semester_3/applied_mechanics_dynamics:bullet-slender-bar-h240"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/bullet-slender-bar-h240
site: BCT Engineering
---

# Bullet embedded in bar, h = 240 mm

Applied Mechanics (Dynamics) · Chapter 6: Plane motion of rigid bodies: energy and momentum methods

- Asked in: 2075 Chaitra, 8 marks
- Video: [Bullet embedded on a panel | Applied Mechanics Dynamic | Plane motion of Rigid Bodies | Unit:9 |PU|](https://www.youtube.com/watch?v=U5xO8ZO4sKk) — Prashant YT
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/bullet-slender-bar-h240
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-6

## Question

A bullet weighing 40gm is fired with a horizontal velocity of 600m/s into the lower end of a slender 7 kg bar of length L = 600mm. Knowing that h = 240mm and that the bar is initially at rest, determine a) the angular velocity of the bar immediately after the bullet becomes embedded. b) The impulsive reaction at C, assuming that the bullet becomes embedded in 0.001s. [Figure: vertical bar AB of length L pivoted at C, distance h below A; bullet with velocity $v_0$ strikes the lower end B.]

## Short answer

Conserve angular momentum about pivot C before and immediately after impact to find ω, then use linear impulse–momentum on the system for the reaction. With h = 240 mm, ω = 35.94 rad/s; the impulsive reaction at C is about 8387 N opposite to the bullet velocity.

## Answer

Angular momentum of the bar and bullet about the pivot $C$ is conserved during the impact (the reaction at $C$ has no moment about $C$; the weights are non-impulsive). Take the bullet mass $m_b = 0.040$ kg, $v_0 = 600$ m/s, bar mass $M = 7$ kg, $L = 0.600$ m, $h = 0.240$ m (distance of the pivot $C$ below $A$); the bullet enters at the lower end $B$.

```text
     A  o
        |   G = centre of bar
   h    |
     C  o-- pivot
        |
     B  o<--- bullet v0
```

#### Distances from C
- Centre of bar $G$ from $C$: $d_G = L/2 - h = 0.300 - 0.240 = 0.060$ m (below $C$)
- Bullet point $B$ from $C$: $d_B = L - h = 0.600 - 0.240 = 0.360$ m

#### Moment of inertia of the bar about C
$$
\bar I = \frac{1}{12}ML^2 = \frac{1}{12}(7)(0.6)^2 = 0.210\ \text{kg m}^2
$$

$$
I_C = \bar I + Md_G^2 = 0.210 + 7(0.060)^2 = 0.2352\ \text{kg m}^2
$$

#### (a) Angular velocity just after impact
The bullet is embedded, so its speed after impact is $\omega d_B$. Conservation of angular momentum about $C$:

$$
m_bv_0d_B = I_C\omega + m_b(\omega d_B)d_B
$$

$$
\omega = \frac{m_bv_0d_B}{I_C + m_bd_B^2} = \frac{0.04\times600\times0.360}{0.2352 + 0.04(0.360)^2} = 35.94\ \text{rad/s}
$$

#### (b) Impulsive reaction at C
Take the bullet direction as positive. Linear impulse-momentum for the system (bar + bullet), the impulse of the reaction being $C\,\Delta t$:

$$
m_bv_0 + C\,\Delta t = M(\omega d_G) + m_b(\omega d_B)
$$

$$
C\,\Delta t = 7(35.94)(0.060) + 0.04(35.94)(0.360) - 0.04(600) = -8.39\ \text{N s}
$$

$$
C = \frac{C\,\Delta t}{\Delta t} = \frac{-8.39}{0.001} = -8387\ \text{N}
$$

The negative sign means the reaction at $C$ acts opposite to the bullet's direction.

**Answer: (a) $\omega = 35.94$ rad/s; (b) $C \approx 8387$ N ($\approx 8.39$ kN) opposite to the bullet velocity.**

## Check yourself

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

1. Which quantity is conserved during the impact, and about what point?
   A. Linear momentum about any point
   B. Angular momentum about the pivot C, since the reaction at C has no moment about C
   C. Kinetic energy
   D. Angular momentum about A
   Answer: B. The unknown reaction at C passes through C and weights are non-impulsive, so moment about C is conserved. (Key point: Conserved quantity)

2. With L = 0.6 m and h = 0.24 m, the distances from C to G and to the bullet point B are:
   A. 0.060 m and 0.360 m
   B. 0.300 m and 0.600 m
   C. 0.240 m and 0.360 m
   D. 0.060 m and 0.600 m
   Answer: A. d_G = L/2 - h = 0.06; d_B = L - h = 0.36. (Key point: Distances)

3. The conservation equation is:
   A. m_b v0 d_B = I_C omega
   B. m_b v0 d_B = I_C omega + m_b (omega d_B) d_B
   C. m_b v0 = (M + m_b) omega
   D. m_b v0 d_B = I_bar omega
   Answer: B. Bullet is embedded, so after impact it adds m_b d_B^2 to the angular momentum: omega = 35.94 rad/s. (Key point: Equation for omega)

4. The impulsive reaction C is found from:
   A. Angular momentum about G
   B. Linear impulse-momentum: m_b v0 + C dt = M omega d_G + m_b omega d_B
   C. Work-energy
   D. C = M g
   Answer: B. Linear impulse-momentum gives C dt = -8.39 N s, so C = -8387 N, opposite to the bullet velocity. (Key point: Impulsive reaction)

## Close variants set in other exams

- [Bullet embedded in bar, h = 260 mm](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Abullet-slender-bar-h260) — Applied Mechanics (Dynamics), chapter 6, asked 2×

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