Answer to Question #281540 in Mechanics | Relativity for Farhad Hassan

Question

A target in a shooting board consists of a vertical square wooden board, 0.250 m on a side and with a mass 0.750 g and pivots around a horizontal axis along its top edge. The board is stuck face-on at its center by a bullet of mass 1.90 g, travelling at 360 m/s that remains embedded in the board.

a) (2 marks) What is the angular speed of the board just after the bullet’s impact?

(b) (2 marks) What maximum height does the center of the board reach from the equilibrium before it starts swinging down again?

(c) (2 marks) What minimum bullet speed is needed for the board to swing all the way over after the impact?

Answer 1

a)

using angular momentum conservation:

$mva/2=(I_1+I_2)\omega$

where m is mass of bullet,

v is speed of bullet,

a is side of board,

$\omega$ is angular speed of the board,

moment of inertia of board:

$I_1=Ma^2/3=0.75\cdot0.25^2/3=0.0156$ kg$\cdot$m²

M is mass of board

moment of inertia of bullet:

$I_2=ma^2/4=0.0019\cdot0.25^2/4=2.97\cdot10^{-5}$ kg$\cdot$m²

then:

$\omega=\frac{mva}{2(I_1+I_2)}=\frac{0.0019\cdot360\cdot0.25}{2(0.0156+2.97\cdot10^{-5})}=5.47$ rad/s

b)

using conservation of energy:

$\frac{(I_1+I_2)\omega^2}{2}=(m+M)gh$

$h=\frac{(I_1+I_2)\omega^2}{2g(m+M)}=\frac{5.47^2(0.0156+2.97\cdot10^{-5}) }{2g(0.0019+0.75)}=0.063$ m

c)

 board will swing all the way over after the impact if its centre reaches highest point

$h=a/2+a/2=a=0.25$ m

then, using conservation of energy:

$\frac{mv^2}{2}=(m+M)gh$

$v=\sqrt{2(m+M)gh/m}=\sqrt{2(0.0019+0.75)\cdot0.25g/0.0019}=44$ m/s