Final answer:
The magnitude and direction of the velocity of the third piece can be determined using the law of conservation of momentum. By setting the initial and final momenta equal to each other, an equation can be derived that relates the velocities of the three pieces. Solving this equation will give the magnitude and direction of the third piece's velocity.
Step-by-step explanation:
The magnitude and direction of the third piece's velocity can be determined using the law of conservation of momentum. Momentum is conserved in an isolated system, such as in this explosion. First, let's calculate the total momentum of the system before the explosion:
Initial momentum = mass of piece 1 * velocity of piece 1 + mass of piece 2 * velocity of piece 2
Since pieces 1 and 2 have the same mass and fly off in different directions with velocities of 50 m/s and 100 m/s, respectively, we can write:
Initial momentum = 2m * 50 m/s + 2m * 100 m/s
where m is the mass of each of the first two pieces.
Next, let's calculate the total momentum of the system after the explosion, taking into consideration the third piece, which has twice the mass of the first two pieces:
Final momentum = mass of piece 1 * final velocity of piece 1 + mass of piece 2 * final velocity of piece 2 + mass of piece 3 * final velocity of piece 3
Since the third piece has twice the mass of the first two pieces, we can write:
Final momentum = 2m * final velocity of piece 1 + 2m * final velocity of piece 2 + 4m * final velocity of piece 3
Since momentum is conserved, the initial momentum and the final momentum are equal:
2m * 50 m/s + 2m * 100 m/s = 2m * final velocity of piece 1 + 2m * final velocity of piece 2 + 4m * final velocity of piece 3
Simplifying this equation, we get:
50 m/s + 100 m/s = final velocity of piece 1 + final velocity of piece 2 + 2 * final velocity of piece 3
Substituting the given values, we have:
50 m/s + 100 m/s = final velocity of piece 1 + final velocity of piece 2 + 2 * final velocity of piece 3
Simplifying further, we get:
150 m/s = final velocity of piece 1 + final velocity of piece 2 + 2 * final velocity of piece 3
Since piece 1 and piece 2 fly off in different directions, their velocities have opposite signs. Let's assume the velocity of piece 1 is positive, and the velocity of piece 2 is negative. Therefore, we can rewrite the equation as:
150 m/s = final velocity of piece 1 - final velocity of piece 2 + 2 * final velocity of piece 3
Now, we also know that the magnitude of the velocity of the third piece is the same as the magnitude of the velocity of the second piece, but in the opposite direction. Let's assume the magnitude of the velocity of the second piece is v. Therefore, the magnitude of the velocity of the third piece is v as well, but in the positive direction. We can rewrite the equation again:
150 m/s = final velocity of piece 1 - (-v) + 2v
Simplifying further, we get:
150 m/s = final velocity of piece 1 + v + 2v
Combining like terms, we have:
150 m/s = final velocity of piece 1 + 3v
Now, we also know that the magnitude of the velocity of the first piece is 50 m/s. Since we assumed the velocity of piece 1 is positive, we can write:
50 m/s = final velocity of piece 1
Substituting this value into the equation, we can solve for v:
150 m/s = 50 m/s + 3v
Simplifying this equation, we get:
100 m/s = 3v
Dividing both sides by 3, we get:
v = 33.33 m/s
Therefore, the magnitude of the velocity of the third piece is 33.33 m/s, and its direction is positive.