Calculating Speed: Lead Sphere & Steel Bolt Collision in Vertical Loop

[zingbang222]

I am having a lot of trouble with this problem. Can anyone help me?

A 20.0 kg sphere of soft lead is suspended from the tip of a pole by a 1.5 m wire such that it is free to swing in a complete verticle circle. Someone takes a crossbow and fires a steel bolt horizontally, mass of 0.50kg, such that it imbeds itself in the soft lead. What was the speed needed by the sphere to perform a complete vertical loop? b) What was the speed initial of the bolt be before the collision?

 

[Andrew Mason]

A 20.0 kg sphere of soft lead is suspended from the tip of a pole by a 1.5 m wire such that it is free to swing in a complete verticle circle. Someone takes a crossbow and fires a steel bolt horizontally, mass of 0.50kg, such that it imbeds itself in the soft lead. What was the speed needed by the sphere to perform a complete vertical loop? b) What was the speed initial of the bolt be before the collision?

The condition for a complete vertical loop is:

\text{Centripetal force} \ge \text{gravitational force}

This occurs when, at the top of the loop:
mv^2/R \ge mg

In order to achieve that, it must have a KE at the bottom of the loop (ie. impact point) that is enough to lift the mass to the top of the circle (ie. one diameter of the circle or 2R) and still have speed v = \sqrt{gR}

So:
\frac{1}{2}mv_0^2 = mg(2R) + \frac{1}{2}mgR
v_0 = \sqrt{5gR}

To do b), you just have to work out the momentum of the bolt needed to give the sphere/bolt this speed.

AM

 

[zingbang222]

Thank you very much. I really appreciate it.