Question 1: Centre of Mass and Rotational Motion

Question

A ball falling freely from a height of $ 4.9 \mathrm{~m} / \mathrm{s} $, hits a horizontal surface. If $ e=\frac{3}{4} $, then the ball will hit the surface, second time after.

Accepted Answer

To determine when the ball will hit the surface for the second time, we start by calculating the velocity just before it hits the surface for the first time. Using the equation for the velocity of a freely falling object, the velocity is:
$  v = \sqrt{2 	imes g 	imes h} = \sqrt{2 	imes 9.8 	imes 4.9} = 9.8 \, 	ext{m/s}  $
where:
$ g = 9.8 \, 	ext{m/s}^2 $ (acceleration due to gravity)
$ h = 4.9 \, 	ext{m} $ (height)
Next, calculate the velocity after the first bounce using the coefficient of restitution $ e = \frac{3}{4} $:
$  v_{	ext{bounce}} = e 	imes 9.8 = \frac{3}{4} 	imes 9.8 = 7.35 \, 	ext{m/s}  $
The time taken for the ball to travel up and back down after the first bounce, until it hits the surface again, is:
$  t = \frac{2v_{	ext{bounce}}}{g} = \frac{2 	imes 7.35}{9.8}  $
$  t = \frac{14.7}{9.8} = 1.5 \, 	ext{s}  $
Therefore, the ball will hit the surface for the second time after 1.5 seconds.

Answer

1.0 s

Answer

1.5 s

Answer

2.0 s

Answer

3.0 s

Centre of Mass and Rotational Motion

A ball falling freely from a height of 4.9 \mathrm{~m} / \mathrm{s} , hits a horizontal surface. If e=\frac{3}{4} , then the ball will hit the surface, second time after.

Choose the correct answer: