A spherical ball weighing 2 kg is dropped from a height of 4.9 m onto an immovable rigid block as shown in the figure. If the collision is perfectly elastic, what is the momentum vector of the ball (in kg m/s) just after impact?

Take the acceleration due to gravity to be 𝑔 = 9.8 m/s² . Options have been rounded off to one decimal place.

The correct option is:
17.0 $\hat{i}$ + 9.8 $\hat{j}$

Step 1: Calculate the velocity of the ball just before impact
The ball of mass m = 2 kg is dropped from a height h = 4.9 m. The magnitude of the velocity $v_0$ just before hitting the inclined plane can be determined using the equations of motion under gravity g = 9.8 m/s²:

$v_0=\sqrt{2gh}$

Substituting the given values:

$v_0=\sqrt{2\times 9.8\times 4.9}=\sqrt{2\times 2\times 4.9\times 4.9}=2\times 4.9=9.8\text{ m/s}$

Since the ball falls vertically downwards in the direction opposite to the unit vector $\hat{j}$, the velocity vector of the ball just before impact is:

${\overrightarrow{v}}_i=-9.8\hat{j}$

Step 2: Determine the geometry of the inclined surface
From the provided image, the rigid block's inclined surface makes an angle of 30° with the horizontal ($\hat{i}$ axis).
A unit tangent vector $\hat{t}$ pointing down along the inclined plane is:

$\hat{t}=cos(30°)\hat{i}-sin(30°)\hat{j}=\frac{\sqrt{3}}{2}\hat{i}-\frac{1}{2}\hat{j}$

The outward unit normal vector $\hat{n}$ perpendicular to this surface (pointing upward and to the right) is:

$\hat{n}=sin(30°)\hat{i}+cos(30°)\hat{j}=\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}$

Step 3: Analyze the elastic collision
Since the collision with the immovable rigid block is perfectly elastic:
1. The component of velocity parallel to the surface (along $\hat{t}$) remains unchanged.
2. The component of velocity perpendicular to the surface (along $\hat{n}$) is reversed in direction.
The final velocity vector ${\overrightarrow{v}}_f$ is given by:

${\overrightarrow{v}}_f={\overrightarrow{v}}_i-2({\overrightarrow{v}}_i·\hat{n})\hat{n}$

First, compute the dot product:

${\overrightarrow{v}}_i·\hat{n}=(-9.8\hat{j})·(\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j})=-9.8\times \frac{\sqrt{3}}{2}=-4.9\sqrt{3}$

Now, calculate the final velocity vector:

${\overrightarrow{v}}_f=-9.8\hat{j}-2(-4.9\sqrt{3})(\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j})$

${\overrightarrow{v}}_f=-9.8\hat{j}+9.8\sqrt{3}(\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j})$

${\overrightarrow{v}}_f=4.9\sqrt{3}\hat{i}+(14.7-9.8)\hat{j}=4.9\sqrt{3}\hat{i}+4.9\hat{j}$

Step 4: Calculate the final momentum vector
The final momentum vector ${\overrightarrow{p}}_f$ is the mass times the final velocity:

${\overrightarrow{p}}_f=m{\overrightarrow{v}}_f=2\times (4.9\sqrt{3}\hat{i}+4.9\hat{j})$

${\overrightarrow{p}}_f=9.8\sqrt{3}\hat{i}+9.8\hat{j}$

Using the approximation $\sqrt{3}\approx 1.732$:

$9.8\times 1.732\approx 16.97\approx 17.0$

Therefore, the final momentum vector of the ball just after impact is:
${\overrightarrow{p}}_f\approx 17.0\hat{i}+9.8\hat{j}\text{ kg m/s}$