A body of mass m moving with a velocity v in the x-direction collides with another body of mass M mo

Q: Wait, why is momentum conserved but not kinetic energy?

Momentum is always conserved in the absence of external forces because it's a direct consequence of Newton's Third Law (action and reaction forces are equal and opposite). Kinetic energy, however, can be transformed into other forms of energy like heat, sound, or deformation energy during the impact, especially when the bodies stick together.

Q: Why do we use vectors for momentum but not for kinetic energy?

Momentum is defined as mass times velocity ($\mathbf{p} = m\mathbf{v}$), and since velocity is a vector, momentum is also a vector. Kinetic energy is defined as $\frac{1}{2}mv^2$, which depends on the square of the speed (a scalar). Energy has no direction!

Q: Can we find the final velocity first instead of using the final momentum for the kinetic energy?

Absolutely! You can find the final velocity vector $\mathbf{v}_f = \frac{mv\hat{\mathbf{i}} + MV\hat{\mathbf{j}}}{m+M}$, find its magnitude squared, and plug it into $K_f = \frac{1}{2}(m+M)v_f^2$. However, using $K_f = \frac{p^2}{2(m+M)}$ is a powerful shortcut that skips a few algebraic steps.

Q: What does 'coalesce' mean physically in this problem?

To 'coalesce' means the two bodies stick together upon impact and move as a single, combined mass of $(m+M)$ afterwards. This is the defining characteristic of a 'perfectly inelastic collision'.

Q: Why did the $m^2v^2$ and $M^2V^2$ terms cancel out in the final fraction calculation?

This is the mathematical manifestation of the energy loss! The terms that cancel out represent the portion of the initial kinetic energy that is retained by the system's center of mass. The remaining cross terms ($mMV^2 + Mmv^2$) represent the internal kinetic energy relative to the center of mass, which is exactly what gets converted into heat.

Question

A body of mass $m$ moving with a velocity $v$ in the $x$-direction collides with another body of mass $M$ moving in the $y$-direction with a velocity $V$. They coalesce into one body during collision. Find
(a) the direction and magnitude of the momentum of the composite body.
(b) the fraction of the initial kinetic energy transformed into heat during the collision.

Sigma Percentile · Accepted Answer

### The Anatomy of a Perfectly Inelastic Collision

Imagine a cosmic intersection. Two celestial bodies are on a collision course. One body, with mass $m$, is cruising along the $x$-axis with a velocity $v$. The other, a more massive entity $M$, is traveling along the $y$-axis with a velocity $V$.

When they meet at the origin, they don't bounce off each other. Instead, they undergo a **perfectly inelastic collision**. They coalesce, fusing together into a single, composite mass of $(m+M)$.

This physical reality sets the stage for our mathematical analysis. In such collisions, while kinetic energy is violently dissipated as heat and deformation, the fundamental law of the universe holds true: **linear momentum is strictly conserved**.

### The Vector Dance of Momentum

Because momentum is a vector quantity, we must treat the $x$ and $y$ directions independently. Before the collision, the total momentum of the system is simply the vector sum of the individual momenta.

The first body contributes momentum in the $x$-direction, giving us $mv\hat{\mathbf{i}}$. The second body contributes in the $y$-direction, giving us $MV\hat{\mathbf{j}}$. Therefore, the initial momentum vector is:
$$ \mathbf{p}_i = mv\hat{\mathbf{i}} + MV\hat{\mathbf{j}} $$

Since momentum is conserved, this must also be the final momentum vector $\mathbf{p}_f$ of the newly formed composite body.

### The Geometry of the Aftermath

To find the magnitude of this final momentum, we turn to the Pythagorean theorem. The total momentum is the hypotenuse of a right-angled triangle formed by the $x$ and $y$ momentum components.
$$ |\mathbf{p}| = \sqrt{(mv)^2 + (MV)^2} $$

But where is this new body heading? We need to find the angle $\alpha$ it makes with the positive $x$-axis. Using simple trigonometry on our momentum triangle, the tangent of this angle is the ratio of the $y$-component to the $x$-component.
$$ 	an\alpha = \frac{p_y}{p_x} = \frac{MV}{mv} $$

Taking the inverse tangent gives us the exact trajectory:
$$ \alpha = 	an^{-1}\left(\frac{MV}{mv}ight) $$

### The Thermodynamics of the Crash

Now, let's address the violent nature of this collision. When the bodies fuse, a significant amount of their initial kinetic energy is transformed into heat. We are tasked with finding the fraction of this lost energy.

The fraction of energy lost is defined as the change in kinetic energy divided by the initial kinetic energy:
$$ 	ext{Fraction} = \frac{K_i - K_f}{K_i} = 1 - \frac{K_f}{K_i} $$

First, we calculate the initial kinetic energy. Unlike momentum, kinetic energy is a scalar, so we simply add the energies of the two bodies:
$$ K_i = \frac{1}{2}mv^2 + \frac{1}{2}MV^2 $$

For the final kinetic energy, we could use the velocity of the composite body, but there is a more elegant path. We can express kinetic energy directly in terms of momentum using the relation $K = \frac{p^2}{2m_{	ext{total}}}$. 
$$ K_f = \frac{|\mathbf{p}|^2}{2(m+M)} = \frac{(mv)^2 + (MV)^2}{2(m+M)} $$

### The Algebraic Symphony

With our energies defined, we substitute them into our fraction formula:
$$ 	ext{Fraction} = 1 - \frac{\frac{(mv)^2 + (MV)^2}{2(m+M)}}{\frac{1}{2}(mv^2 + MV^2)} $$

Notice how the $\frac{1}{2}$ in the numerator and denominator beautifully cancel out. This leaves us with:
$$ 	ext{Fraction} = 1 - \frac{m^2v^2 + M^2V^2}{(m+M)(mv^2 + MV^2)} $$

To combine these terms, we take a common denominator:
$$ 	ext{Fraction} = \frac{(m+M)(mv^2 + MV^2) - (m^2v^2 + M^2V^2)}{(m+M)(mv^2 + MV^2)} $$

Now, we expand the first term in the numerator. It looks intimidating, but watch how the algebra elegantly resolves itself:
$$ (m+M)(mv^2 + MV^2) = m^2v^2 + mMV^2 + Mmv^2 + M^2V^2 $$

When we subtract the $(m^2v^2 + M^2V^2)$ term, the $m^2v^2$ and $M^2V^2$ terms perfectly cancel out! We are left with just the cross terms:
$$ 	ext{Fraction} = \frac{mMV^2 + Mmv^2}{(m+M)(mv^2 + MV^2)} $$

Finally, we factor out the common $Mm$ from the numerator to reveal our pristine final answer:
$$ 	ext{Fraction} = \frac{Mm(v^2 + V^2)}{(m+M)(mv^2 + MV^2)} $$

This elegant equation tells us exactly how much of the universe's motion was sacrificed to the fires of the collision.

Visualized Solution

Initial Setup

Conservation of Linear Momentum

Initial Momentum Vector

Magnitude of Final Momentum

Direction of Final Momentum

Kinetic Energy and Heat

Initial Kinetic Energy

Final Kinetic Energy

Setting up the Fraction

Algebraic Simplification

Taking the Common Denominator

Expanding the Numerator

Final Result

The Sigma Insight: Oblique Collision

The Anatomy of a Perfectly Inelastic Collision

The Vector Dance of Momentum

The Geometry of the Aftermath

The Thermodynamics of the Crash

The Algebraic Symphony

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