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: Analyzing the given information from the question and the diagram
Based on the provided illustration, we can observe the following details:
• Mass of the spherical ball, $m=2\text{kg}$
• Height from which the ball is dropped, $h=4.9\text{m}$
• The acceleration due to gravity, $g=9.8{\text{m/s}}^2$, directed vertically downwards.
• The immovable rigid block has an inclined face that makes an angle of ${30}^{\circ }$ with the horizontal ground.
• The coordinate system is oriented such that $\hat{i}$ points horizontally to the right, and $\hat{j}$ points vertically upwards.

Step 2: Velocity of the ball just before collision
The ball is dropped from rest under the influence of gravity. Using the equation of motion for a freely falling body, the speed $v_0$ just before it hits the inclined surface is:

$$v_0=\sqrt{2gh}$$

Substituting the given values:

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

Since the ball falls vertically downwards in the negative $\hat{j}$ direction, the velocity vector just before collision ${\overrightarrow{v}}_1$ is:

$${\overrightarrow{v}}_1=-9.8\hat{j}\text{m/s}$$

Step 3: Geometry of the inclined surface
Let us define the unit tangent vector $\hat{t}$ along the inclined plane (directed downwards and to the right) and the unit normal vector $\hat{n}$ pointing outwards (upwards and to the right):

$$\hat{t}=\cos \left({30}^{\circ }\right)\hat{i}-\sin \left({30}^{\circ }\right)\hat{j}$$

$$\hat{n}=\sin \left({30}^{\circ }\right)\hat{i}+\cos \left({30}^{\circ }\right)\hat{j}$$

Step 4: Finding the velocity components just before impact
We resolve the pre-impact velocity ${\overrightarrow{v}}_1=-9.8\hat{j}$ into its components along the normal ($v_{1n}$) and tangential ($v_{1t}$) directions:

$$v_{1n}={\overrightarrow{v}}_1\cdot \hat{n}=(-9.8\hat{j})\cdot \left(\sin \right({30}^{\circ })\hat{i}+\cos ({30}^{\circ }\left)\hat{j}\right)=-9.8\cos \left({30}^{\circ }\right)$$

$$v_{1t}={\stackrel{���}{v}}_1\cdot \hat{t}=(-9.8\hat{j})\cdot \left(\cos \right({30}^{\circ })\hat{i}-\sin ({30}^{\circ }\left)\hat{j}\right)=9.8\sin \left({30}^{\circ }\right)$$

Step 5: Applying the conditions of a perfectly elastic collision
Since the collision is perfectly elastic and the rigid block is immovable:
1. The component of velocity parallel (tangential) to the slope remains unchanged:

$$v_{2t}=v_{1t}=9.8\sin \left({30}^{\circ }\right)$$

2. The component of velocity perpendicular (normal) to the slope is reversed:

$$v_{2n}=-v_{1n}=9.8\cos \left({30}^{\circ }\right)$$

Step 6: Velocity vector just after impact
Combining the components back into a single vector ${\overrightarrow{v}}_2$:

$${\overrightarrow{v}}_2=v_{2t}\hat{t}+v_{2n}\hat{n}$$

Substituting the expressions for $\hat{t}$ and $\hat{n}$:

$${\overrightarrow{v}}_2=\left[9.8\sin \right({30}^{\circ }\left)\right]\left[\cos \right({30}^{\circ })\hat{i}-\sin ({30}^{\circ }\left)\hat{j}\right]+\left[9.8\cos \right({30}^{\circ }\left)\right]\left[\sin \right({30}^{\circ })\hat{i}+\cos ({30}^{\circ }\left)\hat{j}\right]$$

Simplifying by factoring out 9.8:

$${\overrightarrow{v}}_2=9.8\left[\right(\sin \left({30}^{\circ }\right)\cos \left({30}^{\circ }\right)+\cos \left({30}^{\circ }\right)\sin \left({30}^{\circ }\right))\hat{i}+({\cos }^2\left({30}^{\circ }\right)-{\sin }^2\left({30}^{\circ }\right)\left)\hat{j}\right]$$

Using trigonometric identities:
• $2\sin \left({30}^{\circ }\right)\cos \left({30}^{\circ }\right)=\sin \left({60}^{\circ }\right)$
• ${\cos }^2\left({30}^{\circ }\right)-{\sin }^2\left({30}^{\circ }\right)=\cos \left({60}^{\circ }\right)$

Substituting these identities:

$${\overrightarrow{v}}_2=9.8\left[\sin \right({60}^{\circ })\hat{i}+\cos ({60}^{\circ }\left)\hat{j}\right]$$

Since $\sin \left({60}^{\circ }\right)=\frac{\sqrt{3}}{2}\approx 0.866$ and $\cos \left({60}^{\circ }\right)=0.5$:

$${\overrightarrow{v}}_2=9.8[0.866\hat{i}+0.5\hat{j}]=8.487\hat{i}+4.9\hat{j}\text{m/s}$$

Step 7: Momentum vector just after impact
The momentum vector $\overrightarrow{P}$ just after collision is given by:

$$\overrightarrow{P}=m{\overrightarrow{v}}_2$$

Substituting $m=2\text{kg}$:

$$\overrightarrow{P}=2\times (8.487\hat{i}+4.9\hat{j})=16.974\hat{i}+9.8\hat{j}\text{kg m/s}$$

Rounding off to one decimal place, we obtain:

$$\overrightarrow{P}\approx 17.0\hat{i}+9.8\hat{j}\text{kg m/s}$$