Question

Q1. A 1.2 g bead slides along a wire, as shown in the figure. At point A, the bead is at rest. Neglect friction. (a) What is the speed of the bead at point B? (b) What is the speed of the bead at point C? 100 cm 80 cm - Potential Energy = 0 В

Answer 1

At point A, Kinetic Energy of Bead =0

And potential Energy of Bead =mgh

Now, m=1.2g =0.0012kg, and h=100cm=1m

So, Total Energy of Bead at A = E_A= 0.01176 J. -----(1)

Now, at point B, potential Energy =0

And kinetic energy = (1/2) mv² = 0.0006 v²

Thus, total energy at B =E_B= 0.0006 v²

Now, using Conservation of Energy, E_A=E_B

So, 0.0006 v² = 0.01176

Thus, v= 4.4272 m/s

Now, again, let the velocity at point C be u.

Then, Kinetic energy at C = (1/2) mu²

and Potential energy at C = mgh=0.8mg =0.009408 J

So, total energy at C= 0.009408 + (1/2) mu²

Using conservation of energy,

Ec=Ea

So, (1/2) mu²=0.01176-0.009408 =0.002352

So, u²=3.92

Hence, u=1.98m/s