Two chunks of ice are sliding on a frictionless frozen pond. Chunk A, with mass mA=5.0kg, moves with initial velocity vA1=2.0m/s parallel to the x-axis. It collides with chunk B, which has a mass mB=3.0kg and is initially at rest. After the collision, the velocity of chunk A is found to be vA2=1.0m/s in a direction making an angle α=30 degrees with an initial direction. What is the final velocity of chunk B?

To find the final velocity of chunk B after the collision, we can use the principle of conservation of linear momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

Before the collision:
The momentum of chunk A before the collision (pA1) is given by:
pA1 = mA * vA1
= 5.0 kg * 2.0 m/s
= 10 kg·m/s

The momentum of chunk B before the collision (pB1) is zero since it is initially at rest.

The total momentum before the collision is the sum of the momenta of chunk A and chunk B:
p_initial = pA1 + pB1
= 10 kg·m/s + 0 kg·m/s
= 10 kg·m/s

After the collision:
The momentum of chunk A after the collision (pA2) is given by:
pA2 = mA * vA2
= 5.0 kg * 1.0 m/s
= 5 kg·m/s

The momentum of chunk B after the collision (pB2) is unknown.

The total momentum after the collision is the sum of the momenta of chunk A and chunk B:
p_final = pA2 + pB2

Using the principle of conservation of linear momentum, we can equate the total momentum before the collision to the total momentum after the collision:
p_initial = p_final
10 kg·m/s = 5 kg·m/s + pB2
pB2 = 10 kg·m/s - 5 kg·m/s
pB2 = 5 kg·m/s

Therefore, the final momentum of chunk B after the collision is 5 kg·m/s. Since momentum is a vector quantity, we can conclude that the final velocity of chunk B is also 5 m/s.

To find the final velocity of chunk B, we can use the principle of conservation of momentum.

The momentum of an object is given by the product of its mass and velocity, p = m * v.

Before the collision:
The momentum of chunk A is: pA1 = mA * vA1 = 5.0 kg * 2.0 m/s = 10.0 kg*m/s.
The momentum of chunk B is initially zero since it is at rest.

After the collision:
Let vA2x and vA2y represent the x and y components of the velocity vector of chunk A after the collision, respectively.

From the given information, we know that the magnitude of the velocity of chunk A, vA2, is 1.0m/s, and it makes an angle α=30 degrees with the initial direction.

Using trigonometry, we can find the x and y components of vA2:
vA2x = vA2 * cos(α) = 1.0 m/s * cos(30°) = 1.0 m/s * 0.866 = 0.866 m/s.
vA2y = vA2 * sin(α) = 1.0 m/s * sin(30°) = 1.0 m/s * 0.5 = 0.5 m/s.

The momentum of chunk A after the collision is:
pA2 = mA * vA2 = 5.0 kg * 1.0 m/s = 5.0 kg*m/s.

According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision:

pA1 + pB1 = pA2 + pB2.

Since chunk B is initially at rest (pB1 = 0), the equation simplifies to:

pA1 = pA2 + pB2.

Substituting the momentum values, we can solve for pB2:

10.0 kg*m/s = 5.0 kg*m/s + pB2.

pB2 = 10.0 kg*m/s - 5.0 kg*m/s = 5.0 kg*m/s.

The final momentum of chunk B is 5.0 kg*m/s. To find the final velocity of chunk B, we can divide this momentum by its mass:

vB2 = pB2 / mB = 5.0 kg*m/s / 3.0 kg ≈ 1.67 m/s.

Therefore, the final velocity of chunk B is approximately 1.67 m/s.