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?

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

The momentum before the collision is equal to the momentum after the collision. Mathematically, this can be written as:

mAvA1 + mBvB1 = mAvA2 + mBvB2

where
mA = mass of chunk A
vA1 = initial velocity of chunk A
mB = mass of chunk B
vB1 = initial velocity of chunk B
vA2 = final velocity of chunk A
vB2 = final velocity of chunk B

In this case, we are given the following values:
mA = 5.0 kg
vA1 = 2.0 m/s
mB = 3.0 kg
vB1 = 0 m/s (since chunk B is initially at rest)
vA2 = 1.0 m/s
α = 30 degrees

First, we need to find the x and y components of the velocity for chunk A after the collision. We can use the given angle α to find these components.

The x-component of velocity for chunk A after the collision can be found using the equation:

vA2x = vA2 * cos(α)

vA2x = 1.0 m/s * cos(30 degrees)
vA2x = 1.0 m/s * 0.866
vA2x = 0.866 m/s

The y-component of velocity for chunk A after the collision can be found using the equation:

vA2y = vA2 * sin(α)

vA2y = 1.0 m/s * sin(30 degrees)
vA2y = 1.0 m/s * 0.5
vA2y = 0.5 m/s

Now, let's substitute these values into the conservation of momentum equation.

(5.0 kg * 2.0 m/s) + (3.0 kg * 0 m/s) = (5.0 kg * 0.866 m/s) + (3.0 kg * vB2)

10.0 kg * m/s = 4.33 kg * m/s + 3.0 kg * vB2

10.0 kg * m/s - 4.33 kg * m/s = 3.0 kg * vB2

5.67 kg * m/s = 3.0 kg * vB2

Dividing both sides by 3.0 kg, we get:

1.89 m/s = vB2

Therefore, the final velocity of chunk B after the collision is 1.89 m/s.

break up chunkA final into x and y components of momentum. You immediately know that chunkB y component of momentum will be be eequal to ChunkA y, but in the opposite direction.

Now, do conservation of momentum in the x direction, and solve for chunkB in the x direction.

knowing the x, y componets of momentum of chunk B, the angle is easily found from the inverse tangent.

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