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GATE 2023 ME – Question 37

Engineering Mechanics · Impulse and momentum, energy formulations · 2 marks · Multiple choice

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/s$^2$. Options have been rounded off to one decimal place.

a 2 kg ball above a rigid triangular block whose top surface is inclined at 30° to the horizontal and slopes down to the right. The axes $\hat{i}$ (horizontal, to the right) and $\hat{j}$ (vertical, up) are shown.
  1. $19.6\,\hat{i}$
  2. $19.6\,\hat{j}$
  3. $17.0\,\hat{i} + 9.8\,\hat{j}$
  4. $9.8\,\hat{i} + 17.0\,\hat{j}$

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Correct answer: (C) $17.0\,\hat{i} + 9.8\,\hat{j}$

Explanation

The speed just before impact is $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 4.9} = 9.8$ m/s, directed downwards, so $\mathbf{v} = (0, -9.8)$. The surface slopes down to the right at 30°, so the unit normal pointing out of the block is $\mathbf{n} = (\sin 30°, \cos 30°) = (0.5, 0.866)$. In a perfectly elastic collision with a fixed surface the normal component of the velocity is reversed: $\mathbf{v}' = \mathbf{v} - 2(\mathbf{v} \cdot \mathbf{n})\mathbf{n}$. Here $\mathbf{v} \cdot \mathbf{n} = -8.487$, so $\mathbf{v}' = (0, -9.8) + 16.97(0.5, 0.866) = (8.49, 4.9)$. The momentum is $2\mathbf{v}' = (16.97, 9.8) \approx 17.0\hat{i} + 9.8\hat{j}$.