Question

Block 1 of mass m1 slides along a frictionless floor and into a one-dimensional elastic collision with stationary block 2 of mass m2 = 8m1. Prior to the collision, the center of mass of the two-block system had a speed of 4.6 m/s.

What is the speed of block 2 after the collision (in m/s)?

Sample submission: 8.9

Answer 1

Mass of the first block = m1

Mass of the second block, m2 = 8*m1

Velocity of center of mass before collision, Vcm = 4.6 m/s

Initial velocity of the first block = v1

Initial velocity of the second block, v2 = 0

Velocity of center of mass, Vcm = ( m1*v1 + m2*v2 ) / ( m1 + m2 )

(m1 + m2 ) Vcm = m1*v1 + m2*v2

(m1 + 8 *m1) * 4.6 = m1 * v1 + m2 * 0

41.4 *m1 = m1 v1

So, v1 = 41.4 m/s

In the case of elastic collision,

Speed of second block after collision is given by

v2' = [ 2 m1 / ( m1 + m2 ) ] * v1 + [ ( m2 - m1 ) / ( m1 + m2 ] * v2

v2' = [ 2 * m1 / ( m1 + 8*m1 ) ] * 41.4 + 0

v2' = ( 2 m1 / 9*m1 ) * 41.4

v2' = 9.2 m/s

Speed of the second block after collision, v2' = 9.2 m/s