A 28g rifle bullet traveling at 430 m/s buries itself in a 3.6kg pendulum hangin

Question

A 28g rifle bullet traveling at 430 m/s buries itself in a 3.6kg pendulum hanging on a 2.8m long string, which makes the pendulum swing upward in an arc. Determine the angle the string makes with the vertical when the pendulum is at its maximum displacement. Show all steps A 28g rifle bullet traveling at 430 m/s buries itself in a 3.6kg pendulum hanging on a 2.8m long string, which makes the pendulum swing upward in an arc. Determine the angle the string makes with the vertical when the pendulum is at its maximum displacement. Show all steps A 28g rifle bullet traveling at 430 m/s buries itself in a 3.6kg pendulum hanging on a 2.8m long string, which makes the pendulum swing upward in an arc. Determine the angle the string makes with the vertical when the pendulum is at its maximum displacement. Show all steps

Explanation / Answer

here,

m1 = 28g = 0.028 kg
v1 = 430 m/s
m2 = 3.6kg
L = 2.8m

From consrvation of energy :
m1v1 + m2v2 = (m1+m2)V
V = m1*v1/(m1+m2)
V = 0.028*430/(0.028+3.6)
V = 3.318 m/s

kinetic Enegy of system will be equal to Potential Energy attained by pendullum :

0.5*(m1+m2)*V^2 = (m1+m2)*g*H
h = V^2/2g
h = 3.318^2/(2*9.8)
h = 0.561 m

y component of disp = 0.561

By using pythgorus theoram :
X = sqrt(2.8^-(2.8-0.561)^2)
x = 0.0757 m

TandA = y/x
A = arcTan(0.561/0.0757)
A = arcTan(7.41)
A = 82.31 degrees

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