Problem 3/247 Page 224, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: Mass m1 impacting mass m2.  Mass m1 has an initial speed of v1.  Mass m2 is stationary.  Coefficient of restitution of e.

Find: The ratio of m1 to m2 such that the final velocity of m2 is greater than v1.

1. Mechanical Systems: System1 is mass m1 during impact.  System 2 is mass m2 during impact.

2. Free Body Diagrams: Not included. System 1 free body diagram shows mass m1 and a large impact force F.  System 2 free body diagram shows mass m2 and the same large impact force F but in the opposite direction from that shown in system 1.

3. Equations:
System 1 Impulse = -Integral of F over impact interval t
System 1 D Linear Momentum = m1 (v1f - v1) t
System 2 Impulse = Integral of F over impact interval t
System 2 D Linear Momentum = m2 (v2f - 0 ) t
e = Speed of separation / Speed of approach = (v2f - v1f) /v1
v2f > v1

4. Solve:
-Integral of F over impact interval = m1 ( v1f - v1)
Integral of F over impact interval =  m2 v2f

Adding the equations:

0 = m1 (v1f - v1) + m2 v2f

Solving for v1f

v1f = v1 - m2/m1 v2f

Plugging this into the coefficient of restitution equation:
e = (v2f - v1 + m2/m1 v2f  ) / v1 = ( 1 + m2/m1 ) v2f/v1 - 1

Solving for v2f:
v2f = (1 + e)/(1+m2/m1) v1

For v2f to be larger than v1, 1+m2/m1 must be less than 1+e, thus:
m2/m1 < e

Inverting
m1/m2 > 1/e