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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.

1. 19.6 î
2. 19.6 ĵ
3. 17.0 î + 9.8 ĵ
4. 9.8 î + 17.0 ĵ

Correct answer: 19.6 ĵ

Solution

The momentum vector just after impact is directed vertically upward, as the ball rebounds after the perfectly elastic collision. The magnitude of the momentum is calculated using the formula p = mv, where the velocity just before impact is determined by the height from which it was dropped, resulting in a momentum of 19.6 kg m/s in the ĵ direction.

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