On a frictionless air track, a blue glider with mass 0.200 kg is moving to the right with a speed of 8.00 m/s. It strikes a red glider that has mass 0.600 kg and that is initially at rest. After the collision, the blue glider is moving to the left with a speed of 2.00 m/s.

(a) What are the magnitude and direction of the velocity of the red glider after the collision?
(b) Is this collision elastic?

a) v = 0.33 to the right.

b) No, it is inelastic

Explanation: mass of blue glider = 0.2kg,

initial velocity of blue glider = 8m/s

Mass of red glider = 0.6 kg

initial velocity of glider = 0 ( since it is at rest)

Final velocity of blue glider after collision = 2 m/s.

Final velocity of the red glider after collision =?

Law of conservation of momentum states vector sum of momentum before collision equals vector sum after collision.

Before collision

(0.2×8) + (0.6×0) = 1.6 + 0 = 1.6 m/s

After collision

they will both move in opposite direction, since blue glider moved left, red glider will move right. Since the blue glider moved in the opposite direction (left), it momentum will be negative and that of red will be positive

-(0.2×2) + 0.6v = 6v - 0.4

By equating them both

1.6 = 6v - 0.4

1.6 + 0.4 = 6v

2 = 6v, v = 2/6

v = 0.33 to the right.

B)

Sum of kinetic energy before collision

1/2 ×(0.2) × 8² + 1/2 × 0.6 × 0

1/2 × 0.2 × 64 = 6.4 J.

Sum of kinetic energy after collision

1/2 × (0.2) ×2² + 1/2 ×(0.6) ×(0.3)²

{1/2 × 0.2 ×4} + 1/2 × 0.6 × 0.09

0.4 + 0.027 = 0.427.

No

Since the kinetic energy before and after collision are not equal, then the collision is inelastic that's the collision is not elastic

unlike poles attract