A gadget of mass 17.85 kg floats in space
without motion. Because of some internal
malfunction, the gadget violently breaks up into 3 fragments flying away from each other.
The first fragment has mass m1 = 4.12 kg
and speed v1 = 3.57 m/s while the second
fragment has mass m2 = 8.02 kg and speed
v2 = 6.03 m/s. The angle between the velocity vectors ~v1 and ~v2 is θ12 = 95
◦
What is the speed v3 of the third fragment?
Answer in units of m/s
resolve the two momentum vectors 1 and 2, then the momentum for piece there is equal and oopposite to that, and you can solve for v3
To find the speed of the third fragment (v3), we can make use of the principle of conservation of momentum. According to this principle, the total momentum before the gadget breaks up should be equal to the total momentum after the breakup.
The momentum of an object is given by the product of its mass and velocity. Let's denote the mass of the third fragment as m3 and its velocity as v3.
Before the breakup, the total momentum is given by the sum of the individual momenta of each fragment:
Total momentum before = momentum of fragment 1 + momentum of fragment 2 + momentum of fragment 3
This can be expressed as:
(m1 * v1) + (m2 * v2) + (m3 * v3) = 0
Since the gadget is initially at rest, the total momentum before is zero.
Now, let's plug in the given values:
(4.12 kg * 3.57 m/s) + (8.02 kg * 6.03 m/s) + (m3 * v3) = 0
Simplifying the equation, we have:
14.6824 kg·m/s + 48.4206 kg·m/s + (m3 * v3) = 0
Combining the two terms on the left side, we get:
63.103 kg·m/s + (m3 * v3) = 0
Since we have the magnitudes of velocity for fragment 1 and fragment 2, we need to use the angle between their velocity vectors (θ12) to determine the direction of the third fragment's velocity. However, since the question doesn't ask for the direction, we can just consider the magnitudes.
To determine the magnitude of the third fragment's velocity, we can use the right triangle formed by vectors v1, v2, and v3 with angle θ12. We can use the law of cosines to express v3 in terms of v1, v2, and θ12:
v3^2 = v1^2 + v2^2 - 2*v1*v2*cos(θ12)
Now we can substitute the given values and solve for v3:
v3^2 = (3.57 m/s)^2 + (6.03 m/s)^2 - 2*(3.57 m/s)*(6.03 m/s)*cos(95°)
v3^2 = 12.7449 m^2/s^2 + 36.3609 m^2/s^2 - 43.2514 m^2/s^2
v3^2 = 5.8544 m^2/s^2
Taking the square root of both sides to find v3:
v3 = √(5.8544 m^2/s^2)
v3 ≈ 2.4184 m/s
Therefore, the speed of the third fragment, v3, is approximately 2.4184 m/s.