The projectile motion of a particle of mass 5 g is shown in the figure. The initial velocity of the

Q: Why is the horizontal velocity constant?

In projectile motion, gravity only acts downwards. Since there is no force acting in the horizontal direction (air resistance is negligible), the horizontal acceleration is zero. Therefore, the horizontal velocity remains constant.

Q: Why does the vertical velocity exactly reverse at point B?

Because point B is at the same height as point A. Using the kinematic equation $v^2 = u^2 + 2as$, the vertical displacement $s$ is zero. This means $v_y^2 = u_y^2$, so $v_y = -u_y$ (since it's moving downwards).

Q: Do we need to calculate the time of flight to solve this?

No, we don't! The beauty of this method is that we only need to compare the initial and final velocity states. The time it takes to get there doesn't affect the final momentum change.

Q: Why did we convert the mass from grams to kilograms?

The final answer is requested in the units of $\text{kg-m/s}$. To ensure our calculation yields the correct numerical value, all inputs must be in standard SI units before multiplying.

Q: What if the angle at point B was different?

On a level horizontal plane without air resistance, the trajectory is perfectly symmetric. The angle of impact will always be equal to the angle of projection. It cannot be different.

Question

The projectile motion of a particle of mass $5	ext{ g}$ is shown in the figure.

The initial velocity of the particle is $5\sqrt{2}	ext{ ms}^{-1}$ and the air resistance is assumed to be negligible. The magnitude of the change in momentum between the points $A$ and $B$ is $x 	imes 10^{-2}	ext{ kg-ms}^{-1}$. The value of $x$ to the nearest integer, is ................. .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a particle launched into the air, tracing a beautiful parabolic path. We are given its mass $m = 5 	ext{ g}$, initial velocity $u = 5\sqrt{2} 	ext{ m/s}$, and the launch angle $	heta = 45^\circ$. Our goal is to find the change in momentum between the launch point $A$ and the landing point $B$.

To find the change in momentum, we need to look at the velocities. Momentum is simply mass times velocity. So, the change in momentum is mass times the change in velocity vector from $A$ to $B$. Mathematically, this is expressed as:

$$\Delta \vec{p} = m(\vec{v}_B - \vec{v}_A)$$

### Breaking Down the Velocities

Let's break down the initial velocity at point $A$ into its horizontal and vertical components. Using basic trigonometry, we get $u \cos 45^\circ$ along the x-axis and $u \sin 45^\circ$ along the y-axis. Therefore, the velocity vector at $A$ is:

$$\vec{v}_A = u \cos 45^\circ \hat{i} + u \sin 45^\circ \hat{j}$$

Now, what happens at point $B$? Since there's no air resistance, the horizontal velocity remains perfectly constant throughout the flight. However, the vertical velocity reverses its direction completely because point $B$ is at the exact same horizontal level as point $A$. So, the vertical component becomes $-u \sin 45^\circ$. The velocity vector at $B$ is:

$$\vec{v}_B = u \cos 45^\circ \hat{i} - u \sin 45^\circ \hat{j}$$

### The Master Equation

Let's calculate the change in velocity. When we subtract the velocity at $A$ from the velocity at $B$, a beautiful thing happens: the horizontal components cancel out completely! We are left with only the vertical components adding up in the negative direction.

$$\Delta \vec{v} = \vec{v}_B - \vec{v}_A = -2u \sin 45^\circ \hat{j}$$

The magnitude of the change in momentum is simply the mass times the magnitude of this change in velocity. The negative sign drops out because we are only interested in the magnitude.

$$|\Delta \vec{p}| = m |\Delta \vec{v}| = 2mu \sin 45^\circ$$

### Final Calculation

Now for the fun part! Let's plug in our given values. Remember to convert the mass from grams to kilograms to keep our units consistent in the SI system. So, $m = 0.005 	ext{ kg}$.

$$|\Delta \vec{p}| = 2 	imes 0.005 	imes 5\sqrt{2} 	imes \frac{1}{\sqrt{2}}$$

The $\sqrt{2}$ terms cancel out perfectly. Multiplying the remaining numbers gives us:

$$|\Delta \vec{p}| = 2 	imes 0.005 	imes 5 = 0.05 	ext{ kg-m/s}$$

This can be written in scientific notation as $5 	imes 10^{-2} 	ext{ kg-m/s}$. Comparing this with the given expression $x 	imes 10^{-2}$, we find that **$x = 5$**!

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Projectile}

\text{Momentum Change Formula}

\text{Velocity at Point A}

\text{Velocity at Point B}

\text{Change in Velocity}

\text{Magnitude of Momentum Change}

\text{Substituting Values}

\text{Final Calculation}

\text{The Way Forward}

The Sigma Insight: Projectile Motion

Analyzing the Setup

Breaking Down the Velocities

The Master Equation

Final Calculation

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