The mass of bob B is 18 times that of bob A. Bob B is initially at rest. If bob A is released from a height of

The mass of bob B is 18 times that of bob A. Bob B is initially at rest. If bob A is released from a height of 1m above its lowest point, what is the maximum height attained by bobs A and B after the collision?

If Ua and Va are the initial and final velocity of A, H 1 and H2 are the initial and final heights of A and Vb is the velocity of B after collision and H3 is the height of B to which it has gone up.

By collision equations,
if m is the mass of A
Va / Ua = (m1 - m2) / (m1 + m2) = - 17/ 19
And
Vb / Ua = 2m1/ (m1+m2) = 2/ 19
--------------------------------------…
Using the formula v² = 2gh,
(H 2/ H 1) = (Va / Ua) ² = (- 17/ 19) ² = 0.8
H 2 = 0.8*1 = 0.8 m
--------------------------------------…
H3/ H1 = (Vb / Ua) ² = (2/ 19) ² = 0.01
H3 =0.01*1 = 0.01 m
======================================…

Derivation of the formula

U refers initial velocity
V refers final velocity
1refers the block 1
2 refers the block 2



Conservation of momentum

m1 U1 = m1 V1 + m2 V2 since the block 2 is at rest U2 = 0

m1 (U1 - V1) = m2 V2. ------------1

Energy is also conserved

m1 (U1² + V1²) = m2 V2²

m1 (U1 + V1)*(U1 – V1) = m2 V2²-----------------2

By equation 1, m1 (U1 - V1) = m2 V2


Hence equation 2 is simplified to
(U1 + V1) = V2---------------------3

From 1
(U1 - V1) = (m2/m1) V2--------------------------4

Adding 3 and 4

2U1 = V2 + (m2/m1) V2 = (m1 + m2) V2 / m1

V2 = 2m1 U1/ (m1 + m2) -------5

Subtracting 3 and 4

2V1 = V2 - (m2/m1) V2

2V1 = (m1 – m2) V2

V1 = (m1 – m2) V2 / 2

V1 = (m1 – m2) U1/ (m1 + m2)

You have to specify whether it's an elastic or inelastic collision......
Elastic:
Ha = .800 m
Hb = .00277 m