Question

L = 1.0 m / 80 = 45° ma = 200 g A - - - - - - - (1) A = 0 (V1A = ( VA mg = 500 g (yo) A = L(1 - cos ) (vo)A = 0 m/s (3)A = L(1 - cos 3) (13)A = 0 m/s (122) A (V2JB A A B (VWB = 0 m/s A 200 [g] steel ball hangs on a 1.0 [m] long string. The ball is pulled sideways so that the string is at a 45° angle, then released. At the very bottom of its swing the ball strikes a steel paperweight which is 500 [g], that is resting on a frictionless table. To what angle does the ball rebound?

Answer 1

Velocity of the ball just before collision is given by (v1x)A = sqrt(2g*y0A)

= sqrt(2*9.8*1*(1-cos 45 degree))

= 2.396 m/s

velocity of the ball after collision (V2x)A = [(m_A - m_B)·v_1A]/(m_A + m_B) = (0.200-0.500)*2.396/[0.200+0.500]

= -1.027 m/s

Let it bound to theta3,

by energy conservation, mg*h = 0.5mv^2

mgL[1- cos theta3] = 0.5mv^2

2gL[1- cos theta3] = v^2

2*9.8*1*(1 - cos theta3) = 1.027^2

(1 - cos theta3) =  1.027^2/19.6 = 0.05381

cos theta3 = 1-0.05381 = 0.9462

theta3 = arccos 0.9462

= 18.9 degree answer