Question

A uniform, 4.5-kg, square, solid wooden gate 1.5 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.1-kg raven flying horizontally at 5.0 m/s flies into this door at its center and bounces back at 2.0 m/s in the opposite direction. (a) What is the angular speed of the gate just after it is struck by the unfortunate raven? (b) During the collision, why is the angular momentum conserved but not the linear momentum?

Answer 1

Answer

given,

mass of the gate = 4.5 kg  

distance between the hinge 1.5 m  

mass of the bird = 1.1 kg

velocity of the bird = 5 m/s

$I_{gate} =\dfrac{1}{3}ML^2$

$I_{gate} =\dfrac{1}{3}\times 4.5 \times 1.5^2$

$I_{gate} =3.375 kg.m^2$

a) from conservation of angular momentum  

$L_i = L_f$  

$mv_1\dfrac{L}{2} = mv_2\dfrac{L}{2} + I_{total} \omega_f$  

$m(v_1+v_2)\dfrac{W}{2}= 3.375 \omega_f$

$1.1(5+2)\dfrac{1.5}{2}= 3.375 \omega_f$

$\omega_f= 1.711\ rad/s$

b) There is no external torque hence, momentum is conserved

    initial kinetic energy = $\dfrac{1}{2}mv^2$

    initial kinetic energy = $\dfrac{1}{2}\times 1.1 \times 5^2$

                                        = 13.75 J

     final kinetic energy

       KE =  $\dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

       KE =  $\dfrac{1}{2}\times 1.1 \times 2^2 + \dfrac{1}{2}\times 3.375\times 1.711^2$

       KE = 7.14 J

Kinetic energy is not conserved

hence, the collision is inelastic.