9. A gun oriented vertically upward fires a bullet of mass 5.00 g with a velocit
ID: 1865719 • Letter: 9
Question
9. A gun oriented vertically upward fires a bullet of mass 5.00 g with a velocity of 900.0 m/s. directly above the gun is a horizontal tree branch of 5.00 cm thickness. The bullet passes through the branch and reaches a maximum height of 1250 m before falling back to earth. (a) Find resistive force exerted by the branch on the bullet, assuming it is constant. (b) If the bullet avoids hitting the branch on the way down find the velocity of the bullet as it returns to the initial position from which it was firedExplanation / Answer
let m = 5 g = 0.005 kg
v1 = 900 m/s
x = 5 cm = 0.05 m
speed of the bulle after passing through the ree branch, v2 = sqrt(2*g*h)
= sqrt(2*9.8*1250)
= 156.5 m/s
a) use Work-energy theorem
Workdone by branch = change in kinetic energy of the bullet
F*x = (1/2)*m*(v2^2 - v1^2)
= (1/2)*m*(v2^2 - v1^2)/x
= (1/2)*0.005*(156.5^2 - 900^2)/0.05
= -3.93*10^4 N
|F| = 3.93*10^4 N <<<<<<<<----------------Answer
b) velocity of the bullet when it reaches the ground, v2 = 156.5 m/s <<<----------------Answer
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