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Chapter 4: Problem 19

With what minimum speed \(v\) must a small ball should be pushed inside a smooth vertical tube from a height \(h\) so that it may reach the top of the tube? Radius of the tube is \(R\). (Assume radius of cross-section of tube is negligible in comparison to \(R\).) (A) \(\sqrt{2 g(h+2 R)}\) (B) \(\frac{5}{2} R\) (C) \(\sqrt{g(5 R-2 h)}\) (D) \(\sqrt{2 g(2 R-h)}\)

Short Answer

Expert verified
The minimum speed $v$ at which the small ball must be pushed inside the vertical tube is \(v =\sqrt{2 g(h+2 R)}\), where $g$ is the acceleration due to gravity, $h$ is the height, and $R$ is the radius of the tube.

Step by step solution

01

Understand the Conservation of Mechanical Energy

In the absence of friction, we can assume that the mechanical energy of the ball is conserved. The mechanical energy of the ball consists of kinetic energy (KE) and potential energy (PE). While initially, it possesses only kinetic energy, at the top of the tube, it has maximum potential energy. The total mechanical energy remains constant.
02

Determine the Initial Kinetic Energy

At the initial position, all the mechanical energy is kinetic energy, given by: \(KE_{initial} = \frac{1}{2} m v^2\), where \(m\) is the mass of the ball and \(v\) is its initial speed.
03

Determine the Potential Energy at Top of the Tube

At the top of the tube, the ball has maximum gravitational potential energy, given by: \(PE_{final} = m g (h+2 R)\), where \(g\) is the acceleration due to gravity.
04

Conservation of Mechanical Energy

As per the conservation of mechanical energy, the total mechanical energy at the initial position and the top of the tube must be equal. \(KE_{initial} = PE_{final}\) \[\frac{1}{2} m v^2 = m g (h+2 R)\]
05

Solve for the Initial Speed

Now, let's solve the equation for the initial speed \(v\): \[\frac{1}{2} v^2 = g (h+2 R)\] \[v^2 = 2 g (h+2 R)\] \[v = \sqrt{2 g(h+2 R)}\] Hence, the correct answer is option (A), \(v =\sqrt{2 g(h+2 R)}\).

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