Question

A bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length script l and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? (Use any variable or symbol stated above along with the following as necessary: g. Be sure to use script l from physPad.)

v =

Answer 1

`v = √(16\cdot g \cdot L)`

Step-by-step explanation:

The physical phenomenon is described by the Principles of Momentum Conservation and Energy Conservation:

Momentum

`m \cdot v = M\cdot (v)/(2) + m \cdot v'`

Energy

`(1)/(2)\cdot m \cdot v^(2) = (1)/(8)\cdot M \cdot v^(2) + (1)/(2)\cdot m \cdot v'^(2)`

`(1)/(8)\cdot M\cdot v^(2) = 2\cdot M\cdot g \cdot L`

The minimum speed of the pendulum bob so that it could barely swing through a complete vertical cycle is:

`(1)/(8)\cdot v^(2) = 2\cdot g\cdot L`

`v^(2) = 16\cdot g\cdot L`

`v = √(16\cdot g \cdot L)`