Question

A 0.400 kg green bead slides on a curved frictionless wire, starting from rest at point A in Figure P6.52. At point B the bead collides elastically with a 0.540 kgblue ball at rest. Find the maximum height the blue ball rises as it moves up the wire.

Answer 1

Energy at point A = mgh a
= 0.4 kg * 9.8 m / s 2 *5 m
= 19.6 J
From law of conservation of energy energy at point B = energy at point A
( 1/ 2) m v 2 = 19.6 J
from this velocity of the bead at point B v =v[ 2*19.6 / m ]
= 9.89 m / s


now,
at point B elastic collision.
so v1- v2= u2 -u1
or v1- v2 = 0 - 9.89 therefore v2 = v1 + 9.89 -----> (1)

by conservation of momentum we have
m1v1 + m2v2 = m1u1 + m2u2
hence
0.4v1 + 0.54v2 = 0.4*9.89 + 0.54 *0 ----> (2)

solving (1) and (2) you get values of v1 and v2

then use 1/2 m (v2)^2 = m2 gh to find h