A cue ball (mass = 0.150 kg) is at rest on a frictionless pool table. The ball is hit dead center by a pool stick which applies an impulse of +1.45 N·s to the ball. The ball then slides along the table and makes an elastic head-on collision with a second ball of equal mass that is initially at rest. Find the velocity of the second ball just after it is struck?

1.45/0.15

1.45/0.15 = 9.66 m/s

To find the velocity of the second ball just after it is struck, we can use the principle of conservation of linear momentum.

The principle of conservation of linear momentum states that in the absence of external forces, the total linear momentum of a system remains constant.

In this case, since the collision is elastic, momentum is conserved before and after the collision. The initial momentum for the system (cue ball + second ball) is zero, as both balls are initially at rest.

Let's define the variables:
- Mass of the cue ball (m₁) = 0.150 kg
- Mass of the second ball (m₂) = 0.150 kg
- Initial velocity of the cue ball (v₁) = 0 (since it's at rest)
- Impulse applied to the cue ball (J) = +1.45 N·s

Using the formula for impulse:

J = Δp
1.45 N·s = (m₁ * v₁) + (m₂ * v₂)

Since the cue ball is at rest initially, its initial momentum (m₁ * v₁) is zero. Therefore, we can simplify the equation to:

1.45 N·s = m₂ * v₂

Now, we can solve for the velocity of the second ball (v₂):

v₂ = (1.45 N·s) / m₂

Plugging in the values:

v₂ = (1.45 N·s) / 0.150 kg
v₂ ≈ 9.67 m/s

Therefore, the velocity of the second ball just after it is struck is approximately 9.67 m/s.