A 0.150-kg glider is moving to the right with a speed of 0.80 m/s on a frictionless, horizontal air track. The glider has a head-on collision with a 0.300-kg glider that is moving to the left with a speed of 2.20 m/s. Find the final velocity (magnitude and direction) of each glider if the collision is elastic.

The final speed of 0.150-kg glider after collision is 3.2 m/s to the left

The final speed of 0.300-kg glider after collision is 0.20 m/s to the left

Explanation:

Given;

mass of glider moving to the right, m₁ = 0.150-kg

mass of glider moving to the left, m₂ = 0.300-kg

initial speed of glider moving to the right, u₁ = 0.80 m/s

initial speed of glider moving to the left, u₂ = 2.20 m/s

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.15 x 0.8) + (0.3 x - 2.2) = 0.15v₁ + 0.3v₂

-0.54 = 0.15v₁ + 0.3v₂

0.15v₁ + 0.3v₂ = - 0.54 equation (i)

Again, their relative velocity after collision is given as;

u₁ - u₂ = v₂ - v₁

0.8 - (-2.2) = v₂ - v₁

3 = v₂ - v₁

v₂ =  v₁ + 3  equation (ii)

Substitute v₂ in equation (i)

0.15v₁ + 0.3v₂ = - 0.54

0.15v₁ + 0.3(v₁ + 3 ) = - 0.54

0.15v₁ + 0.3v₁ + 0.9 = - 0.54

0.45v₁  = - 0.54 - 0.9  

0.45v₁  = -1.44

v₁ = -1.44 /0.45

v₁ = - 3.2 m/s

Thus, the final speed of 0.150-kg glider after collision is 3.2 m/s to the left

From equation (ii), v₂ = v₁ + 3

v₂ = -3.2 + 3

v₂ = - 0.20 m/s

Therefore, the final speed of 0.300-kg glider after collision is 0.20 m/s to the left

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