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 = 5m₁. Prior to the collision, the center of mass of the two-block system had a speed of 8.40 m/s.

a) what is the speed of the center of mass after the collision?

b)What is the speed if the block 2 after the collisions?

Answer 1

Velocity of the center of mass of the two particle before collision = 8.4 m/s.
Take the velocity of mass m1 before collision = v1 m/s
Velocity of the mass m2 = 5m1 before collision = 0
velocity of center of mass = (m1 x v1 + m2 x 0) / (m1+m2) = 8.4 m/s.
(m1 x v1) / 6 m1 = 8.4
v1 = 8.4 x 6 = 50.4 m/s.

Take the velocity of mass m1 after collision = v2.
Take the velocity of mass m2 after collision = v3.
Linear momentum is always conserved,
Momentum before collision = momentum after collision
m1 v1 = m1 v2 + m2 v3

a)

Speed of center of mass after collision = Speed of center of mass before collision.
=( m1 v2 + m2 v3) / (m1 + m2)
= (m1 v1) / (m1 + 5m2)
​= v1/6 = 8.4 m/s

b)

​Substituting m2 = 5 m1 in the above equation,
​v1 = v2 + 5 v3 ...(1)
Since the momentum is elastic, kinetic energy is conserved,
1/2 m1 (v1)^​2 = 1/2 m1 (v2)² + 1/2 m2 (v3)²,
Substituting m2 = 5 m1,
(v1)² = (v2)² + 5 (v3)²   ...(2)
Substituting v2 = v1 - 5 v3 in equation (2),
we get, v3 = 1/3 v1
v3 = 16.8 m/s.

  
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