Question

A 0.00410–kg bullet traveling horizontally with a speed of 1.00 ✕ 10³ m/s enters a 21.0–kg door, embedding itself 19.0 cm from the side opposite the hinges as in the figure below. The 1.00–m–wide door is free to swing on its hinges.

(a) Before it hits the door, does the bullet have angular momentum relative to the door's axis of rotation?

Yes or No     

Explain.

(b) Is mechanical energy conserved in this collision? Answer without doing a calculation.

Yes or No     

(c) At what angular speed does the door swing open immediately after the collision? (The door has the same moment of inertia as a rod with axis at one end.)
rad/s

(d) Calculate the energy of the door–bullet system. (Enter the kinetic energy of the door–bullet system just after collision.)
J

Determine whether the energy of the door–bullet system is less than or equal to the kinetic energy of the bullet before the collision.

less than or equal  

Answer 1

Let
m = 0.0041 kg
v = 1*10^3 m/s
M = 21 kg
L = 1 m
r = 1 - 0.19

= 0.81 m

a) Yes.

angular momentum of the bullet relative to the door's axis of rotation = r cross p

= r*p*sin(90)

= r*m*v

= m*v*r

b) No
Because the collision is completely inelastic collision.

c) let w is the angular speed of door just after bullet embeded into the door.

Apply conservation of angular momentum

final angular momentum = initial angular momentum

I*w = m*v*r

(M*L^2/3 + m*r^2)*w = m*v*r

w = m*v*r/(M*L^2/3 + m*r^2)

= 0.004*1*10^3*0.81/(21*1^2/3 + 0.004*0.81^2)

= 0.463 rad/s <<<<<<<<---------Answer

d) the energy of the door–bullet system, KE = (1/2)*I*w^2

= (1/2)*(M*L^2/3 + m*r^2)*w^2

= (1/2)*(21*1^2/3 + 0.004*0.81^2)*0.463^2

= 0.750 J <<<<<<<<<-------------Answer

e) Lessthan

Because, in a completley inelastic collision final kinetic energy is always less than initial kinetic energy.