9. + -/6 points SerPSE9 11.P.041 My Notes A 0.00400-kg bullet traveling horizont

Question

9. + -/6 points SerPSE9 11.P.041 My Notes A 0.00400-kg bullet traveling horizontally with speed 1.00 x 103 m/s strikes a 17.1-kg door, embedding itself 11.4 cm from the side opposite the hinges as shown in the figure below. The 1.00-m wide door is free to swing on its frictionless hinges. Hinge (a) Before it hits the door, does the bullet have angular momentum relative the door's axis of rotation? O No (b) If so, evaluate this angular momentum. (If not, enter zero.) kg m2/s If not, explain why there is no angular momentumm This answer has not been graded yet. (c) Is mechanical energy of the bullet-door system constant in this collision? Answer without doing a calculation. Yes O No (d) At what angular speed does the door swing open immediately after the collision?

Explanation / Answer

here,

mass of bullet , mb = 0.004 kg

initial speed , u = 1 * 10^3 m/s

mass of door , md = 17.1 kg

x = 0.114 m

(A)

before it hits the ground , YES, the bullet have angular momentum relative the door's axis of rotation

(B)

the angular momentum , L = m * u * ( 1 - x)

L = 0.004 * 1000 * ( 1 - 0.114)

L = 3.54 kg.m^2/s

the angular momentum is 3.54 kg.m^2/s

(C)

No, the mechanical energy is not conserved in this collison,

as mehanical energy is conserved when the collison is perfectly elastic

but this collison is an inelastic collison

(D)

let the angular speed be w

using conservation of angular momentum

mb * u * ( 1 - x) = ( Id + Ib) * w

0.004*1000*(1-0.114) = ( 17.1*1^2 + 0.004*(1-0.114)^2) * w

w = 0.21 rad/s

the angular speed of gate is 0.21 rad/s

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