Question

A 0.006 00-kg bullet traveling horizontally with a speed of 1.00 103 m/s enters an 19.1-kg door, imbedding itself 8.20 cm from the side opposite the hinges as in the figure below. The 1.00-m wide door is free to swing on its frictionless hinges.



(a) Before it hits the door, does the bullet have angular momentum relative the door's axis of rotation?
Yes
No


(b) If so, evaluate this angular momentum. (If not, enter zero.)

__________m^2/s


(c) Is mechanical energy of the bullet-door system constant in this collision? Answer without doing a calculation.
Yes
No


(d) At what angular speed does the door swing open immediately after the collision?

______________rad/s


(e) Calculate the energy of the bullet-door system and determine whether it is less than or equal to the kinetic energy of the bullet before the collision.
KEf =

_________J

KEi =

_________J

I know that a is yes and c is no, but I'm not sure how to do the others.

Answer 1

(a) Here the bullet travels with constant velocity at perpendicular distance \(d\) from axis of rotation. Thus the bullet has angular momentum about door's axis of rotation.

(b) Length at which the bullet hits the door from hinge is,

$$ \begin{aligned} d &=1.0 \mathrm{~m}-10.0 \mathrm{~cm} \\ &=1.0 \mathrm{~m}-(8.20 \mathrm{~cm})\left(\frac{1 \mathrm{~m}}{100 \mathrm{~cm}}\right) \\ &=0.918 \mathrm{~m} \end{aligned} $$

Initial angular momentum of the bullet is, \(L=m v d\)

$$ \begin{array}{l} =(0.006 \mathrm{~kg})\left(1.00 \times 10^{2} \mathrm{~m} / \mathrm{s}\right)(0.918 \mathrm{~m}) \\ =5.508 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s} \end{array} $$

(c) Here the collision is inelastic, since the bullet gets embedded into the door. So the mechanical energy is not conserved.

(d) Moment of inertia of the bullet and the door after the butlet embedded in door

$$ \begin{aligned} I &=I_{d}+m d^{2} \\ &=\frac{1}{3} M L^{2}+m d^{2} \\ &=\frac{1}{3}(19.1 \mathrm{~kg})(1.0 \mathrm{~m})^{2}+(0.006 \mathrm{~kg})(0.918 \mathrm{~m})^{2} \\ &=6.3717 \mathrm{~kg} \cdot \mathrm{m}^{2} \end{aligned} $$

Initial angular momentum of the bullet \(L_{3}=m v d\) Apply law of conservation of angular momentum just be fore and after collision to the bullet and door system.

$$ \begin{aligned} L_{i} &=L_{f} \\ m^{d} &=I \omega \\ \omega &=\frac{m v d}{I} \\ &=\frac{(0.006 \mathrm{~kg})\left(1.00 \times 10^{3} \mathrm{~m} / \mathrm{s}\right)(0.918 \mathrm{~m})}{6.3717 \mathrm{~kg} \cdot \mathrm{m}^{2}} \\ &=0.864 \mathrm{rad} / \mathrm{s} \end{aligned} $$

(e) Expression for the rotational kinetic energy

$$ \begin{aligned} K E_{f} &=\frac{1}{2} I \omega^{2} \\ &=\frac{1}{2}\left(63717 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)(0.864 \mathrm{rad} / \mathrm{s})^{2} \\ &=2.78 \mathrm{~J} \end{aligned} $$

Initial kinetic energy of the bullet is

$$ \begin{aligned} K E_{\mathrm{i}} &=\frac{1}{2} m^{2} \\ &=\frac{1}{2}(0.006 \mathrm{~kg})\left(1.00 \times 10^{3} \mathrm{~m} / \mathrm{s}\right)^{2} \\ &=3000 \mathrm{~J} \end{aligned} $$

From the above cal culations we say that the rotational kine tic energy is less than the kinetic energy of the bullet.