A 0.450-kg ice puck, moving east with a speed of 5.34 m/s , has a head-on collision with a 0.990-kg puck initially at rest. Assume that the collision is perfectly elastic.1. What is the speed of the 0.450-kg puck after the collision?2. What is the direction of the velocity of the 0.450-kg puck after the collision?3. What ise the speed of the 0.990-kg puck after the collision?4. What is the direction of the velocity of the 0.990-kg puck after the collision?

(1) -2.0025m/s.

(2) It moves west.

(3) 3.3375m/s.

(4) It moves east.

Assuming the collision occurred in an isolated system. This means that total momentum of the system of pucks is conserved. Since no external forces are acting on these pucks, the momentum of the pucks before collision is equal to the momentum of the pucks after collision. i.e

(p₁)₀ + (p₂)₀ = (p₁)₁ + (p₂)₁              -------------(i)

Where;

(p₁)₀ = momentum of the 0.45kg puck before collision

(p₂)₀ = momentum of the 0.990kg puck before collision

(p₁)₁ = momentum of the 0.45kg puck after collision

(p₂)₁ = momentum of the 0.990kg puck after collision

But;

(p₁)₀ = m₁ u₁  

[m₁ = mass of the 0.45kg, u₁ = speed of the 0.45kg before collision]

(p₂)₀ = m₂u₂

[m₂ = mass of the 0.990kg, u₂ = speed of the 0.990kg before collision]

(p₁)₁ = m₁v₁

[m₁ = mass of the 0.45kg, v₁ = speed of the 0.45kg after collision]

(p₂)₁ = m₂v₂

[m₂ = mass of the 0.990kg, v₂ = speed of the 0.990kg after collision]

Equation (i) then becomes;

m₁ u₁ + m₂u₂ = m₁v₁ + m₂v₂    ----------------(ii)

From the question:

m₁ = 0.450kg

u₁ = +5.34m/s   [Taking east direction as positive]

m₂ = 0.990kg

u₂ = 0m/s    [since the second puck is initially at rest]

Substitute these values into equation (ii)

(0.450 x 5.34) + (0.990 x 0) = 0.45 v₁ + 0.990 v₂

2.403 + 0 = 0.45 v₁ + 0.990 v₂

2.403 = 0.45 v₁ + 0.990 v₂           ------------------(iii)

Also, since the collision is perfectly elastic, it means that the kinetic energy is conserved. i.e the total kinetic energy before collision is equal to the total kinetic energy after collision.

=> m₁ u²₁ + m₂u²₂ = m₁v²₁ + m₂v²₂

Substitute the necessary values into the above equation:

[ x 0.45 x 5.34²] + [0] = [ x 0.45 x v²₁] + [ x 0.990 x v²₂]

[6.41601] = [0.225 x v²₁] + [0.495 x v²₂]  ------------------(iv)

Now let's solve equations (iii) and (iv) simultaneously

2.403 = 0.45 v₁ + 0.990 v₂

6.41601 = 0.225 x v²₁ + 0.495 x v²₂

let

v₁ = x

v₂ = y

2.403 = 0.45 x + 0.990 y                  ------------(5)

6.41601 = 0.225 x² + 0.495 y²         -------------(6)

From equation (5), make x subject of the formula

2.403 = 0.45x + 0.990y

0.45x = 2.403 - 0.990y         [divide through by 0.45]

x = 5.34 - 2.2y    ----------------(m)

Substitute x into equation (6)

6.41601 = 0.225 (5.34 - 2.2y)² + 0.495 y²     [expand bracket]

6.41601 = 0.225 [28.5156 - 23.496y + 4.84y²] + 0.495 y²   [remove bracket]

6.41601 = 6.41601 - 5.2866y + 1.089y² + 0.495 y²

1.584y² - 5.2866y = 0

y(1.584y - 5.2866) = 0

y = 0       or       1.584y - 5.2866 = 0

y = 0       or       1.584y = 5.2866

y = 0       or       y = 3.3375

Since y = v₂ cannot be zero because the puck will definitely move after collision, the second value of y = 3.3375 is considered.

Substitute this value into equation (m)

x = 5.34 - 2.2y

x = 5.34 - 2.2(3.3375)

x = 5.34 - 7.3425

x = -2.0025

Therefore,

x = v₁ = -2.0025m/s

y = v₂ = 3.3375m/s

(1) From the analyses above, the speed of the 0.450kg puck after collision is -2.0025m/s.

(2) Since the speed is negative, it shows that the 0.45kg puck moves opposite the direction at which it was moving before collision. It moves west.

(3) The speed of the 0.990kg puck after collision is 3.3375m/s.

(4) Since the speed is positive, it shows that the 0.990kg puck moves east. Remember that east has been taking as the positive direction.

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