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Chapter 5: Problem 60

A gun fires a shell and recoils horizontally. If the shell travels along the barrel with speed \(v\), the ratio of speeds with which the gun recoils, if the barrel is (i) horizontal (ii) inclined at an angle of \(30^{\circ}\) with horizontal, is (A) 1 (B) \(\frac{2}{\sqrt{3}}\) (C) \(\frac{\sqrt{3}}{2}\) (D) \(\frac{1}{2}\)

Short Answer

Expert verified
The ratio of the speeds with which the gun recoils when the barrel is horizontal and when it's inclined is \(2/\sqrt{3}\), so the correct option is (B) \(2/\sqrt{3}\).

Step by step solution

01

Calculate horizontal recoil speed when the barrel is horizontal

Firstly, let's denote the mass of the gun as \(M\) and the mass of the shell as \(m\). With the barrel horizontal, the initial momentum is zero as everything is at rest. When the gun is fired, the shell is projected with a speed \(v\). Therefore, by conservation of momentum, \(Mv_{1} = mv\), where \(v1\) represents the recoil speed of the gun.This gives us \(v_{1} = \frac{m}{M}v\).
02

Calculate horizontal recoil speed when the barrel is inclined at 30 degrees

When the barrel is inclined, the shell follows a different path and hence has a horizontal velocity \(v\cos(30)\), and by conservation of momentum, the horizontal recoil speed of the gun, \(v_{2}\) is given by \(Mv_{2} = mv\cos(30)\). This simplifies to \(v_{2} = \frac{m}{M}v\cos(30)\).
03

Get the ratio of the recoil speeds

The ratio \(v_{1}/v_{2}\) of the recoil speeds is hence \(\frac{m}{M}v / (\frac{m}{M}v\cos(30)) = 1/ \cos(30) = 2/\sqrt{3}\).

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