A block moving horizontally on a smooth surface with a speed of 40 m/s splits into two parts with ma

Q: Why is momentum conserved but not kinetic energy?

During an explosion or splitting event, internal forces do work, which converts internal potential energy (like chemical or elastic energy) into kinetic energy. However, since these internal forces cancel each other out, the net external force is zero, keeping momentum conserved.

Q: How did we get the masses $m_1$ and $m_2$ as $M/3$ and $2M/3$?

The problem states the masses are in a 1:2 ratio. This means the total mass $M$ is divided into $1 + 2 = 3$ equal parts. The smaller piece gets 1 part ($M/3$), and the larger piece gets 2 parts ($2M/3$).

Q: What does 'fractional change' mean?

Fractional change is the change in a quantity divided by its initial value. It tells you how much the quantity changed relative to what it started with. Formula: $(Final - Initial) / Initial$.

Q: Could the fragments move in opposite directions?

Yes, depending on the internal forces. However, the conservation of momentum equation would naturally yield a negative velocity for one of the fragments if it moved backwards. Here, both velocities are positive, meaning they move in the same direction.

Q: Why did the kinetic energy increase?

The block split due to some internal mechanism (like an explosion or spring release). This mechanism did positive work on the fragments, adding kinetic energy to the system.

Question

A block moving horizontally on a smooth surface with a speed of $40	ext{ m/s}$ splits into two parts with masses in the ratio of $1:2$. If the smaller part moves at $60	ext{ m/s}$ in the same direction, then the fractional change in kinetic energy is

Options:

(A) $1/3$

(B) $2/3$

(C) $1/8$

(D) $1/4$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a block of mass $M$ moving on a smooth horizontal surface. It is cruising along at a steady speed of $40 	ext{ m/s}$. Suddenly, the block splits into two fragments: a smaller piece $m_1$ and a larger piece $m_2$.

The problem states that their masses are in a $1:2$ ratio. Let us express these masses in terms of the original mass $M$. Since the ratio is $1:2$, the total parts are $3$. Therefore, the mass of the smaller fragment is $m_1 = \frac{M}{3}$, and the mass of the larger fragment is $m_2 = \frac{2M}{3}$.

### The Master Equation
Since there are no external horizontal forces acting on the system during the split, we can confidently apply the principle of conservation of linear momentum. The initial momentum of the system must equal the final momentum.

$$p_i = p_f$$

$$Mv = m_1v_1 + m_2v_2$$

Let us substitute our known values into the equation. The initial momentum is $M 	imes 40$. For the final momentum, we plug in $\frac{M}{3}$ for $m_1$ and $60$ for $v_1$:

$$M(40) = \left(\frac{M}{3}ight)(60) + \left(\frac{2M}{3}ight)v_2$$

Notice how the mass $M$ cancels out from both sides. Solving this simple linear equation gives us the velocity of the larger block, $v_2$:

$$40 = 20 + \frac{2}{3}v_2$$

$$20 = \frac{2}{3}v_2 \implies v_2 = 30 	ext{ m/s}$$

### Calculating the Energies
Our ultimate goal is to find the fractional change in kinetic energy. This is defined as the final kinetic energy minus the initial kinetic energy, all divided by the initial kinetic energy:

$$	ext{Fractional Change} = \frac{K_f - K_i}{K_i} = \frac{K_f}{K_i} - 1$$

First, let us calculate the initial kinetic energy $K_i$. Using $\frac{1}{2}Mv^2$ and plugging in $40$ for $v$, we get:

$$K_i = \frac{1}{2}M(40)^2 = 800M$$

Now for the final kinetic energy $K_f$. This will be the sum of the kinetic energies of both fragments. Let us carefully substitute the masses and velocities we found:

$$K_f = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$$

$$K_f = \frac{1}{2}\left(\frac{M}{3}ight)(60)^2 + \frac{1}{2}\left(\frac{2M}{3}ight)(30)^2$$

Let us compute this. The first term simplifies to $600M$, and the second term becomes $300M$. Adding them up, the total final kinetic energy is:

$$K_f = 600M + 300M = 900M$$

### The Final Reveal
Finally, let us plug these into our fractional change formula:

$$\frac{\Delta K}{K_i} = \frac{900M}{800M} - 1$$

This simplifies to $\frac{9}{8} - 1$, giving us our final answer:

$$\frac{\Delta K}{K_i} = \frac{1}{8}$$

**Notice something interesting?** The final kinetic energy is actually greater than the initial. Where did this extra energy come from? The internal energy (perhaps chemical or elastic) that caused the block to split was converted into this additional kinetic energy!

Animated Solution for Physics - System of Particles: A block moving horizontally on a smooth surface with a speed of $40 m/s$ splits into two parts with masses in the ratio of $1 : 2$ . If the smaller part moves at $60 m/s$ in the same direction, then the fractional change in kinetic energy is

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Visualized Solution

$I ni t ia l S t a t e$

$F ina l S t a t e$

$M a sseso f F r a g m e n t s$

$C o n ser v a t i o n o f M o m e n t u m$

$S u b s t i t u t in g V a l u es$

$S o l v in g f or v_{2}$

$F r a c t i o na l C han g e in K in e t i c E n er g y$

$I ni t ia l K in e t i c E n er g y (K_{i})$

$F ina l K in e t i c E n er g y (K_{f})$

$C a l c u l a t in g K_{f}$

$F ina l C a l c u l a t i o n$

$T h e W a y F or w a r d$

The Sigma Insight: Conservation of Linear Momentum

Analyzing the Setup Imagine a block of mass $M$ moving on a smooth horizontal surface

The Master Equation Since there are no external horizontal forces acting on the system during the split, we can confidently apply the principle of conservation of linear momentum

Calculating the Energies Our ultimate goal is to find the fractional change in kinetic energy

The Final Reveal

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Visualized Solution

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