Answer to Question #281895 in Physics for Fahad

Question

3. A target in a shooting board consists of a vertical square wooden board, 0.250 m on a side and with a mass 750 g and pivots around a horizontal axis along its top edge. The board is stuck faceon 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

Given:

M is the mass of the square;

m is the mass of the bullet;

v is the speed of the bullet;

a is the side of the square;

I₁ is the moment of inertia of the square;

I₂ is the moment of inertia of the bullet.

Solution:

(a) Apply conservation of angular momentum:

$$\frac{mva}2=(I_1+I_2)\omega,\\\space\\ \frac{mva}2=\bigg(\frac{Ma^2}3+\frac{ma^2}{4}\bigg)\omega,\\\space\\ \omega=\frac{6mv}{a(4M+3m)}=5.5\text{ rad/s}.$$

(b) Apply conservation of energy:

$$\frac12(I_1+I_2)\omega^2=(m+M)gH,\\\space\\ \frac12\bigg(\frac{Ma^2}3+\frac{ma^2}{4}\bigg)\omega^2=(m+M)gH,\\\space\\ H=\frac{(a\omega)^2(4M+3m)}{24g(M+m)}=0.032\text{ m}.$$

(c) The speed is

$$\frac12mv^2=(m+M)gH,\\\space\\ v=\sqrt{2gh(M+m)/m}=44 \text{ m/s}.$$