Answer to Question #1183 in Mechanics | Relativity for Jon Allen
69.6 kg. <br>The bullet emerges with a speed v/2. <br>The pendulum bob is suspended by a stiff rod
of length 0.449 m and negligible mass. <br>The acceleration of gravity is 9.8 m/s2 .
<br><br>What is the minimum value of v such that the pendulum bob will barely swing through
a complete vertical circle? <br>Answer in units of m/s.
mv = mv/2 + MV;
MV2/2 = 2MgL.
Here m - is the mass of the bullet, M - the mass of the pendulum, v - speed of the bullet, V - the speed of the pendulum after interaction, L - is the length of the stiff.
V = 2√(gL)
mv/2 = 2M√(gL)
v = 4M/m √(gL) = 4*69.6/0.00114*√(9.8*0.449) = 512,271.964 m/s
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