A hockey puck moving due east with speed 1.00 m/s collides with an identical puck moving 40 degrees south of west with speed 2.00 m/s. After they collide, the first puck is moving south with speed 1.50 m/s. What is the direction and speed of the second puck?

break up each puck into N, E components. Then use the law of conservation of momentum in each direction.

N:
0-M*2sin40=-M*1.5+M*Xn
notice the M's divide out. Solve for Xn, the north component of the second puck.
E:
M*1-2Mcos40=0+MXe
solve for Xe, the east component of the second puck.

To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy.

Let's break down the velocities into their x and y components.

The initial velocity of the first puck moving due east can be represented as (1.00 m/s, 0 m/s) since it has no y-component velocity.

The initial velocity of the second puck moving 40 degrees south of west can be broken down into its x and y components as follows:
vx = 2.00 m/s * cos(40°)
vy = -2.00 m/s * sin(40°)

The resulting velocity of the first puck after the collision can be represented as (0 m/s, -1.50 m/s) since it is moving south.

Now, let's denote the final velocity of the second puck after the collision as (Vx, Vy), where Vx is the x-component and Vy is the y-component of its velocity.

According to the conservation of momentum, the sum of the x-components and y-components of momentum before the collision should be equal to the sum of the x-components and y-components of momentum after the collision.

In the x-direction:
Initial momentum = (1.00 m/s) * (mass of the first puck) + (2.00 m/s * cos(40°)) * (mass of the second puck)
Final momentum = Vx * (mass of the first puck) + Vx * (mass of the second puck)

In the y-direction:
Initial momentum = 0 * (mass of the first puck) + (-2.00 m/s * sin(40°)) * (mass of the second puck)
Final momentum = -1.50 m/s * (mass of the first puck) + Vy * (mass of the second puck)

Since the two pucks are identical, their masses cancel out in both momentum equations.

By using the conservation of kinetic energy, we know that the initial kinetic energy equals the final kinetic energy after the collision.

(1/2) * m * (1.00 m/s)^2 + (1/2) * m * (2.00 m/s)^2 = (1/2) * m * (0 m/s)^2 + (1/2) * m * (-1.50 m/s)^2 + (1/2) * m * Vx^2 + (1/2) * m * Vy^2

Simplifying this equation, we get:
1 + 4 = 2.25 + (1/2) * (Vx^2 + Vy^2)
5 = 2.25 + (1/2) * (Vx^2 + Vy^2)

Now, let's solve for Vx and Vy.

Simplifying further:
Vx^2 + Vy^2 = 5 - 2.25
Vx^2 + Vy^2 = 2.75

Since the two pucks are identical, Vx = Vy.

Substituting this into the equation:
2 * Vx^2 = 2.75
Vx^2 = 2.75 / 2
Vx^2 = 1.375
Vx ≈ √1.375
Vx ≈ 1.17 m/s

Therefore, the direction and speed of the second puck after the collision is approximately 1.17 m/s, at an angle of approximately 45 degrees south of east.