Question

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.

Hint:

When dealing with collisions on an inclined plane, resolving the velocity vector into components parallel and perpendicular to the plane simplifies the problem immensely. Remember: for a smooth surface, the parallel component is unchanged, and for a perfectly elastic collision, the perpendicular component just reverses its sign.

Answer 1

Step 1: Understanding the Concept:
This is a problem of perfectly elastic collision with an inclined plane. We need to find the velocity of the ball just before impact, then resolve this velocity into components parallel and perpendicular to the inclined surface. For a perfectly elastic collision with a fixed surface, the velocity component perpendicular to the surface is reversed, while the component parallel to the surface remains unchanged. Finally, we convert the post-impact velocity vector back to the global \(\mathbf{\hat{i}}, \mathbf{\hat{j}}\) coordinate system and calculate the momentum.
Step 2: Key Formula or Approach:
1. Calculate the velocity \(v_i\) just before impact using conservation of energy: \(mgh = \frac{1}{2}mv_i^2\).
2. Define a local coordinate system (\(n, t\)) with \(n\) normal to the incline and \(t\) tangent to the incline.
3. Resolve the incoming velocity vector \(\mathbf{v}_i\) into components \(v_{in}\) and \(v_{it}\).
4. Apply the rules for perfectly elastic collision: \(v_{fn} = -v_{in}\) and \(v_{ft} = v_{it}\).
5. Combine the final components \(v_{fn}\) and \(v_{ft}\) to get the final velocity vector \(\mathbf{v}_f\) in the local system.
6. Convert \(\mathbf{v}_f\) back to the global \(\mathbf{\hat{i}}, \mathbf{\hat{j}}\) system. 7. Calculate the final momentum \(\mathbf{p}_f = m\mathbf{v}_f\).
Step 3: Detailed Calculation:
1. Velocity before impact: The ball is dropped from rest. \[ v_i = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 4.9} = \sqrt{96.04} = 9.8 \text{ m/s} \] The velocity vector just before impact is \(\mathbf{v}_i = -9.8 \mathbf{\hat{j}}\).
2. Resolve velocity into components along the incline:
The incline is at \(30^\circ\) to the horizontal. The normal to the incline is at \(60^\circ\) to the vertical (\(\mathbf{\hat{j}}\) axis) and \(30^\circ\) to the horizontal (\(\mathbf{\hat{i}}\) axis).
The angle between the incoming velocity vector (\(-\mathbf{\hat{j}}\)) and the normal direction is \(60^\circ\).
- Component perpendicular (normal) to the surface: \(v_{in} = v_i \cos(60^\circ) = 9.8 \times 0.5 = 4.9\) m/s (directed into the surface).
- Component parallel (tangential) to the surface: \(v_{it} = v_i \sin(60^\circ) = 9.8 \times \frac{\sqrt{3}}{2} \approx 8.487\) m/s (directed down the incline).
3. Velocity components after impact:
- \(v_{fn} = -v_{in} = -4.9\) m/s (directed away from the surface).
- \(v_{ft} = v_{it} = 8.487\) m/s (still directed down the incline).
4. Convert final velocity back to \(\mathbf{\hat{i}}, \mathbf{\hat{j}}\) coordinates:
The tangential velocity vector \(\mathbf{v}_{ft}\) is along the incline (at \(30^\circ\) below horizontal). \[ \mathbf{v}_{ft} = 8.487 (\cos(30^\circ)\mathbf{\hat{i}} - \sin(30^\circ)\mathbf{\hat{j}}) = 8.487(0.866\mathbf{\hat{i}} - 0.5\mathbf{\hat{j}}) = 7.35\mathbf{\hat{i}} - 4.24\mathbf{\hat{j}} \] The normal velocity vector \(\mathbf{v}_{fn}\) is perpendicular to the incline (at \(60^\circ\) above horizontal). \[ \mathbf{v}_{fn} = 4.9 (\cos(60^\circ)\mathbf{\hat{i}} + \sin(60^\circ)\mathbf{\hat{j}}) = 4.9(0.5\mathbf{\hat{i}} + 0.866\mathbf{\hat{j}}) = 2.45\mathbf{\hat{i}} + 4.24\mathbf{\hat{j}} \] The total final velocity is \(\mathbf{v}_f = \mathbf{v}_{ft} + \mathbf{v}_{fn}\): \[ \mathbf{v}_f = (7.35 + 2.45)\mathbf{\hat{i}} + (-4.24 + 4.24)\mathbf{\hat{j}} \] Correction in angle decomposition: Let's use angles with respect to the horizontal (\(+\mathbf{\hat{i}}\) axis).
- Incoming velocity \(\mathbf{v}_i\) is at \(-90^\circ\).
- The normal vector \(\mathbf{n}\) is at \(90^\circ+30^\circ = 120^\circ\).
- The tangent vector \(\mathbf{t}\) is at \(30^\circ\).
- \(\mathbf{v}_i = 9.8\text{ at } -90^\circ\).
- \(\mathbf{v}_{it} = (\mathbf{v}_i \cdot \mathbf{t})\mathbf{t} = (v_i \cos(\theta_{it})) \mathbf{t}\). Angle between \(-90^\circ\) and \(30^\circ\) is \(120^\circ\), but the direction is important.
Let's use a simpler angle approach. The incline is at an angle \(\theta=30^\circ\) to the horizontal.
- Velocity before impact: \(\mathbf{v}_i = (0, -v_i)\) where \(v_i = 9.8\).
- Velocity after impact \(\mathbf{v}_f = (v_{fx}, v_{fy})\).
The reflection law for velocity states \(v_f = v_i\) and the angle of reflection equals the angle of incidence with respect to the normal.
The angle of incidence is \(60^\circ\). So the angle of reflection is also \(60^\circ\). The incoming vector makes an angle of \(-90^\circ\) with the horizontal. The normal makes an angle of \(120^\circ\). The reflected ray will make an angle of \(120^\circ + (120^\circ - (-90^\circ)) = 120^\circ + 210^\circ = 330^\circ\), which is wrong. The angle of the final velocity vector relative to the normal is \(180^\circ\) minus the angle of the incident velocity vector relative to the normal. This is too complex. Let's stick to components.
Re-evaluating components:
Let's define the tangent direction as \(t\) (down the slope) and normal as \(n\) (out of the slope).
- Tangent vector: \(\mathbf{t} = \cos(-30^\circ) \mathbf{\hat{i}} + \sin(-30^\circ) \mathbf{\hat{j}} = \frac{\sqrt{3}}{2}\mathbf{\hat{i}} - \frac{1}{2}\mathbf{\hat{j}}\).
- Normal vector: \(\mathbf{n} = \cos(60^\circ) \mathbf{\hat{i}} + \sin(60^\circ) \mathbf{\hat{j}} = \frac{1}{2}\mathbf{\hat{i}} + \frac{\sqrt{3}}{2}\mathbf{\hat{j}}\).
- Incoming velocity: \(\mathbf{v}_i = -9.8 \mathbf{\hat{j}}\).
- \(v_{it} = \mathbf{v}_i \cdot \mathbf{t} = (-9.8 \mathbf{\hat{j}}) \cdot (\frac{\sqrt{3}}{2}\mathbf{\hat{i}} - \frac{1}{2}\mathbf{\hat{j}}) = (-9.8)(-\frac{1}{2}) = 4.9\). So \(\mathbf{v}_{it} = 4.9\mathbf{t}\).
- \(v_{in} = \mathbf{v}_i \cdot \mathbf{n} = (-9.8 \mathbf{\hat{j}}) \cdot (\frac{1}{2}\mathbf{\hat{i}} + \frac{\sqrt{3}}{2}\mathbf{\hat{j}}) = (-9.8)(\frac{\sqrt{3}}{2}) = -4.9\sqrt{3}\). So \(\mathbf{v}_{in} = -4.9\sqrt{3}\mathbf{n}\).
- After collision: \(\mathbf{v}_{ft} = \mathbf{v}_{it} = 4.9\mathbf{t}\) and \(\mathbf{v}_{fn} = -\mathbf{v}_{in} = 4.9\sqrt{3}\mathbf{n}\).
- Final velocity: \(\mathbf{v}_f = 4.9\mathbf{t} + 4.9\sqrt{3}\mathbf{n}\).
\[ \mathbf{v}_f = 4.9\left(\frac{\sqrt{3}}{2}\mathbf{\hat{i}} - \frac{1}{2}\mathbf{\hat{j}}\right) + 4.9\sqrt{3}\left(\frac{1}{2}\mathbf{\hat{i}} + \frac{\sqrt{3}}{2}\mathbf{\hat{j}}\right) \] \[ \mathbf{v}_f = \left(\frac{4.9\sqrt{3}}{2} + \frac{4.9\sqrt{3}}{2}\right)\mathbf{\hat{i}} + \left(-\frac{4.9}{2} + \frac{4.9\sqrt{3}\sqrt{3}}{2}\right)\mathbf{\hat{j}} \] \[ \mathbf{v}_f = 4.9\sqrt{3} \mathbf{\hat{i}} + \left(-\frac{4.9}{2} + \frac{4.9 \times 3}{2}\right)\mathbf{\hat{j}} = 4.9\sqrt{3} \mathbf{\hat{i}} + \left(\frac{-4.9 + 14.7}{2}\right)\mathbf{\hat{j}} \] \[ \mathbf{v}_f = 4.9\sqrt{3} \mathbf{\hat{i}} + \frac{9.8}{2} \mathbf{\hat{j}} = 4.9\sqrt{3} \mathbf{\hat{i}} + 4.9 \mathbf{\hat{j}} \] \[ \mathbf{v}_f \approx 8.487 \mathbf{\hat{i}} + 4.9 \mathbf{\hat{j}} \] 5. Calculate final momentum: \(m=2\) kg. \[ \mathbf{p}_f = m\mathbf{v}_f = 2 \times (8.487 \mathbf{\hat{i}} + 4.9 \mathbf{\hat{j}}) = 16.974 \mathbf{\hat{i}} + 9.8 \mathbf{\hat{j}} \] 6. Round off to one decimal place: \[ \mathbf{p}_f \approx 17.0 \mathbf{\hat{i}} + 9.8 \mathbf{\hat{j}} \] Step 4: Final Answer:
The final momentum vector is \(17.0 \mathbf{\hat{i}} + 9.8 \mathbf{\hat{j}}\).
Step 5: Why This is Correct:
The calculation correctly determines the impact velocity and then uses the principle of elastic collision with a fixed surface: the tangential velocity component is conserved, and the normal velocity component is reversed. The vector decomposition into and recomposition from the surface-aligned coordinate system is performed correctly.