Question

Three small balls of the same size but different masses are hung side-by-side in parallel on the strings of same length. They touch each other. The first ball of mass m1 is pulled to the height h and released from rest. The first ball collides elastically with the second ball of mass m2, and the second ball collides elastically with the third ball of mass m3 afterwards. After these two collisions, three balls will have the same momentum. What is the value of m3?
(a) m1
(b) 2m1/3
(c) m1/6
(d) m1/2
(e) m1/3

Answer 1

Answer:

m1/6 ( c )

Explanation:

since all the balls starts having the same momentum after the two collisions we will apply the principal of conservation of energy

After first collision

m1v = m1v1 + m2v2 --- ( 1 )

After second collision

m2v2 = m2v2 + m3v3   ---- ( 2 )

combining equations 1 and 2

m1v = m1v1 + m2v2 + m3v3  ----- ( 3 )

All balls moving at the same momentum ( p ) = m1v1 = m2v2 = m3v3

note ; 3p = m1v ∴ m3 = $\frac{m1v}{3v3}$  -----  ( 4 )

applying conservation of energy

3v = v1 + v2 + v3 ------- ( 5 )

also 3m1v1 = m1v = v1 = v/3 =

v2 + v3 = 8/3 v ----- ( 6 )

next eliminate V3 for equation 6 by applying conservation of energy and momentum

m1 =  2m2 ------ ( 7 )

now using p1 = p2 = m1v1 = 1/2 m1v1  hence v2 = 2v1  where v1 = 1/3 v

hence ; v2 = 2/3 v ------- ( 8 )

solving with equation 6 and 8

v3 = 2v ------ ( 9 ) ∴  v/v3 = 1/2 ---- ( 10 )

solving with equation 9 and 10

m3 = m1/3 * 1/2 = m1/6