@Frank Apisa,
Frank Apisa wrote:Why not tackle Brandon's question, DNA...or acknowledge that you cannot...
Maybe he's having a problem with this specific question, rather than being completely ignorant of basic physics. I guess he was absent from school on the day that they studied the conservation of momentum.
The problem was:
A bullet of mass 10 grams srikes a simple pendulum of mass 2 kg and remains embedded in it. The center of mass of the pendulum rises a vertical distance of 12 cm. Calculate the bullet's initial speed.
Let Mb = mass of the bullet
Mp = mass of the pendulum
Vi = velocity of the bullet before the collision
Vf = velocity of pendulum with embedded bullet just after the collions
The momentum of the bullet before the collision is equal to the momentum of the pendulum with a bullet embedded after the collision.
Mb Vi = (Mb + Mp) Vf
By conservation of energy:
½ (Mb + Mp) Vf^2 = (Mb + Mp)gh
Vf ^2 = 2gh
Vf = SQRT(2gh)
Plugging this back in to the above relation between Vi and Vf:
Vi = [(Mb + Mp)/Mb] SQRT(2gh)
Plugging in Mb = .01 kg, Mp = 2 kg, h = .12m, and g = 9.8 m/s^2, this gives:
(2.01 kg/.01 kg) SQRT[2 ( 9.8 m/s^2) (.12m)] = 308.3 m/s
But here is your chance to redeem yourself and show that you really are qualified to correct the world scientific community:
A circular curve of road is designed for traffic moving at 40 mi/hr. If the radius of the road’s curve is 400 ft, what is the correct angle of banking for the road?