- #1

- 81

- 0

**Hi,**

A rod of mass 'm' and length 'l' is lying on a smooth horizontal surface. A particle of mass 'm' is moving with a velocity [tex]v_{0}[/tex] strikes the rod at the centre and sticks to it. What is the velocity of end A after impact? [Solution: [tex]\displaystyle \frac{v_{0}}{2}[/tex]]

http://img131.imageshack.us/img131/2803/angmom1si.png [Broken][/URL]

A rod of mass 'm' and length 'l' is lying on a smooth horizontal surface. A particle of mass 'm' is moving with a velocity [tex]v_{0}[/tex] strikes the rod at the centre and sticks to it. What is the velocity of end A after impact? [Solution: [tex]\displaystyle \frac{v_{0}}{2}[/tex]]

http://img131.imageshack.us/img131/2803/angmom1si.png [Broken]

**Here is what I did:**

Considering both the particle and the rod as the system, no external torque acts. Therefore, angular momentum can be conserved about any point.

Conserving angular momentum about A,

[tex]\displaystyle mv_{0}\frac{l}{2} = I_{A}\omega[/tex]

[tex]\displaystyle mv_{0}\frac{l}{2} = (\frac{ml^2}{3} + {\frac{ml}{4}}^2)\omega[/tex]

[tex]\displaystyle \omega = \frac{6v_{0}}{7l}[/tex]

Velocity of end A = [tex]\displaystyle \frac{\omega}{(\frac{l}{2})} = \frac{12v_{0}}{7l^2}[/tex]

What am I doing wrong?

Considering both the particle and the rod as the system, no external torque acts. Therefore, angular momentum can be conserved about any point.

Conserving angular momentum about A,

[tex]\displaystyle mv_{0}\frac{l}{2} = I_{A}\omega[/tex]

[tex]\displaystyle mv_{0}\frac{l}{2} = (\frac{ml^2}{3} + {\frac{ml}{4}}^2)\omega[/tex]

[tex]\displaystyle \omega = \frac{6v_{0}}{7l}[/tex]

Velocity of end A = [tex]\displaystyle \frac{\omega}{(\frac{l}{2})} = \frac{12v_{0}}{7l^2}[/tex]

What am I doing wrong?

Last edited by a moderator: