Question

A 0.350kg bead slides on a curved fritionless wire,

Answer 1

Answer:

h2 = 0.092m

Explanation:

From a balance of energy from point A to point B, we get speed before the collision:

$m1*g*h-\frac{m1*V_B^2}{2}=0$  Solving for Vb:

$V_B=\sqrt{2gh}=6.56658m/s$

Since the collision is elastic, we now that velocity of bead 1 after the collision is given by:

$V_{B'}=V_B*\frac{m1-m2}{m1+m2} = \sqrt{2gh}* \frac{m1-m2}{m1+m2}=-1.34316m/s$

Now, by doing another balance of energy from the instant after the collision, to the point where bead 1 stops, we get the distance it rises:

$m1*g*h2-\frac{m1*V_{B'}^2}{2}=0$ Solving for h2:

h2 = 0.092m