Part A A target in a shooting gallery consists of a vertical square wooden board

Question

Part A A target in a shooting gallery consists of a vertical square wooden board, 0.230 m on a side and with mass 0.730 kg that pivots on a horizontal axis along its top edge. The board What is the angular speed of the board just after the bullet's impact? is struck face-on at its center by a bullet with mass 1.70 g that is traveling at 305 m/s and that remains embedded in the board. rad/s Submit My Answers Give Up Part B What maximum height above the equilibrium position does the center of the board reach before starting to swing down again? Im

Explanation / Answer

=====================

length of the wooden board, l=0.23 m

mass of the board, m1=0.73 kg

mass of the bullet, m2=1.7g

speed of the bulllet, v=305 m/sec

A)

by using conservation angular mometum,

m2*V*l/2=(m2*(l/2)^2+m1*l^2/3)*w

1.7*10^-3*305*(0.23/2)=(1.7*10^-3*(0.23/2)^2+0.73*(0.23^2)/3)*w

==> w=4.62 rad/sec

amgular speed, w=4.62 rad/sec

B)

by using law of consrvation enegry,

1/2*I*w^2=(m1+m2)*g*h

1/2*(m2*(l/2)^2+(m1*l^2/3)*w^2=(m1+m2)*g*h

1/2*(1.7*10^-3*(0.23/2)^2+(0.73*0.23^2/3)*(4.62)^2=(0.73+1.7*10^-3)*9.8*h

=====> h=19.16 mm

C)

if h=l

use

1/2*(1.7*10^-3*(0.23/2)^2+(0.73*0.23^2/3)*(w')^2=(0.73+1.7*10^-3)*9.8*19.11*10^-3

====> W'=21.25 rad/sec

d)

m2*V*l/2=(m2*(l/2)^2+m1*l^2/3)*w'

1.7*10^-3*v*(0.23/2)=(1.7*10^-3*(0.23/2)^2+0.73*(0.23^2)/3)*(21.25)

===> v=1401.6 m/sec

speed of the bullert, v=1401.6 m/sec

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.