A block of mass 1.9 kg is at rest at the edge of a table of height 1 m. A bullet of mass 0.1 kg coll

Q: Why can't we just use conservation of mechanical energy from the very beginning?

Because the collision between the bullet and the block is perfectly inelastic! During this violent impact, a significant amount of kinetic energy is lost to heat, sound, and deformation. Mechanical energy is only conserved *after* the collision is over and the combined mass is falling under the influence of gravity alone.

Q: Why is linear momentum conserved during the collision if gravity is acting on the system?

Gravity acts in the vertical direction, but the collision happens entirely in the horizontal direction. Since there are no external horizontal forces (like friction on the table), the horizontal momentum is perfectly conserved.

Q: Do I need to calculate the final velocity of the block right before it hits the ground?

You could, using kinematics, but it's completely unnecessary! The beauty of the work-energy theorem is that it allows us to bypass complex kinematic equations. The total mechanical energy at the top of the table is exactly equal to the kinetic energy at the bottom.

Q: What happens to the kinetic energy that was lost during the collision?

It doesn't just disappear from the universe! It is converted into non-mechanical forms of energy. Specifically, it goes into the thermal energy (heating up the bullet and the wood), acoustic energy (the "bang" sound), and the work done to deform the wood as the bullet embeds itself.

Q: Why did the problem explicitly state to "assume there is no rotational motion"?

If the bullet hit the block off-center, it would exert a torque, causing the block to spin as it falls. This would mean some of the potential energy would be converted into rotational kinetic energy ($\frac{1}{2}I\omega^2$), leaving less for translational kinetic energy. By ignoring rotation, we keep the math focused purely on translational motion.

Question

A block of mass 1.9 kg is at rest at the edge of a table of height 1 m. A bullet of mass 0.1 kg collides with the block and sticks to it. If the velocity of the bullet is 20 m/s in the horizontal direction just before the collision, then the kinetic energy just before the combined system strikes the floor, is (Take, $g = 10 	ext{ m/s}^2$ and assume there is no rotational motion and loss of energy after the collision is negligible)

Options:

(A) 20 J

(B) 19 J

(C) 21 J

(D) 23 J

Sigma Percentile · Accepted Answer

This problem is a beautiful combination of two fundamental principles of physics: the conservation of linear momentum and the conservation of mechanical energy. It requires us to carefully analyze the physical situation and break it down into two distinct phases. Let's dive into the story of the speeding bullet and the resting block!

### Phase 1: The Inelastic Collision

Imagine a wooden block of mass $M = 1.9 	ext{ kg}$ resting peacefully at the edge of a table. Suddenly, a bullet of mass $m = 0.1 	ext{ kg}$ comes flying in horizontally with a velocity of $u = 20 	ext{ m/s}$.

When the bullet strikes the block, it doesn't bounce off; it embeds itself deep within the wood. This is the hallmark of a **perfectly inelastic collision**. During this violent impact, a significant amount of kinetic energy is lost—converted into heat, sound, and the work done to deform the wood. Therefore, we **cannot** use the conservation of mechanical energy for this phase.

However, because the collision happens entirely in the horizontal direction and there are no external horizontal forces acting on the system (like friction from the table), the **linear momentum** of the system is perfectly conserved.

Let's set up our master equation for momentum:

$$p_{	ext{initial}} = p_{	ext{final}}$$

The initial momentum is just the bullet moving, while the final momentum is the combined mass $(m + M)$ moving together with a new common velocity, $v$.

$$m u = (m + M) v$$

Now, we substitute our known values into the raw setup:

$$0.1 	imes 20 = (0.1 + 1.9) v$$

$$2 = 2v$$

$$v = 1 	ext{ m/s}$$

So, immediately after the collision, the combined system of the block and the bullet slides off the edge of the table at a speed of $1 	ext{ m/s}$.

### Phase 2: The Free Fall and Energy Conservation

Now begins the second phase of our story. The combined mass is in free fall, dropping from a height of $h = 1 	ext{ m}$.

As it falls, gravity does work on the system, pulling it downward and accelerating it. Because gravity is a conservative force and we are ignoring air resistance, the **total mechanical energy** of the system is now conserved. The energy it has at the top of the table will be exactly equal to the energy it has just before it strikes the floor.

Let's define our energy states. At the top of the table, the system has both kinetic energy (because it's moving horizontally at $1 	ext{ m/s}$) and gravitational potential energy (because it's $1 	ext{ m}$ above the ground).

$$E_{	ext{total}} = K_{	ext{initial}} + U_{	ext{initial}}$$

$$E_{	ext{total}} = \frac{1}{2}(m+M)v^2 + (m+M)gh$$

Just before the system strikes the floor, its height is zero, meaning all of its potential energy has been converted into kinetic energy. Therefore, the total energy we calculate at the top will be exactly equal to the final kinetic energy $K_{	ext{final}}$ at the bottom!

### The Final Calculation

Let's plug our numbers into the energy equation. The combined mass is $m + M = 0.1 + 1.9 = 2 	ext{ kg}$.

$$K_{	ext{final}} = \frac{1}{2}(2)(1)^2 + (2)(10)(1)$$

$$K_{	ext{final}} = 1 + 20$$

$$K_{	ext{final}} = 21 	ext{ J}$$

And there we have it! The kinetic energy of the combined system just before it strikes the floor is **$21 	ext{ J}$**.

This problem elegantly demonstrates why we must be careful about *when* to apply conservation laws. Momentum got us through the messy collision, and energy conservation carried us smoothly to the ground.

Select Answer:

Visualized Solution

$m = 0.1 kg, M = 1.9 kg$

$p_{ini t ia l} = p_{f ina l}$

$m u = (m + M) v$

$v = 1 m/s$

$E_{t o t a l} = K_{f ina l}$

$K_{f ina l} = \frac{1}{2} (m + M) v^{2} + (m + M) g h$

$K_{f ina l} = \frac{1}{2} (2) (1)^{2} + (2) (10) (1)$

$K_{f ina l} = 21 J$

$Elastic vs Inelastic Collision$

The Sigma Insight: Conservation of Mechanical Energy

Phase 1

Phase 2

The Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m=0.1 kg,M=1.9 kg

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } pinitial​=pfinal​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } mu=(m+M)v

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=1 m/s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Etotal​=Kfinal​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kfinal​=21​(m+M)v2+(m+M)gh

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kfinal​=21​(2)(1)2+(2)(10)(1)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kfinal​=21 J

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Elastic vs Inelastic Collision