A particle of mass `0.1kg` is launched at an angle of `53^(@)` with the horizontal . The particle enters a fixed rough hollow tube whose length is slightly less than `12.5m` and which is inclined at an angle of `37^(@)` with the horizontal as shown in figure. It is known that the velocity of ball when it enters the tube is parallel to the axis of the tube. The coefficient of friction betweent the particle and tube inside the tube is `mu=(3)/(8)[` Take `g=10m//g^(2)]`

The kinetic energy of the particle when it comes out of the tube is approximately equal to `:`

Options

A. zero
B. `4J`
C. `7.2J`
D. `11.2J`

Answer

A

Explanation

The initial velocity is `15m//s` acceleration of practicle is `(g sin theta + mu g cos theta )`
downwards along the tube.

`:. A=10 sin 37 +(3)/(8) xx10 xx cos 37=10xx(3)/(5)+(3)/(8)xx 10 xx(4)/(5)=9m//s^(2)`
` :. V ^(2)=u^(2)+2as`
`V^(2)=15^(2)-2xx9xx12.5=0`
`:. V^(2)=0`
Given that the tube is slightly less than `12.5m`. It means the particle will just drop from tube. Hence `K.E.` at the tube end `=0`