A 3.0kg object originally at rest explodes and splits into three fragments. One fragment has a mass of .50kg and flies off along the negative x-axis at a speed of 2.8m/s, and another has a mass of 1.3kg and flies off along the negative y-axis at a speed of 1.5m/s. What are the speed and direction of the third fragment?
467
To find the speed and direction of the third fragment, we need to apply the principle of conservation of momentum. According to this principle, the total momentum before the explosion is equal to the total momentum after the explosion.
Let's break down the given information:
- Initial object mass (m1) = 3.0 kg
- Fragment 1 mass (m2) = 0.50 kg
- Fragment 1 velocity (v2) = -2.8 m/s (negative x-direction)
- Fragment 2 mass (m3) = 1.3 kg
- Fragment 2 velocity (v3) = -1.5 m/s (negative y-direction)
We can find the momentum before the explosion by multiplying the initial object mass by its initial velocity (which is 0 since the object is initially at rest):
Initial momentum (before explosion) = m1 * v1
The momentum after the explosion can be calculated by summing the individual momenta of the three fragments:
Momentum (after explosion) = (m2 * v2) + (m3 * v3) + (m4 * v4)
Now, equating the initial and final momentum, we get:
m1 * v1 = (m2 * v2) + (m3 * v3) + (m4 * v4)
Substituting the given values:
3.0 kg * 0 m/s = (0.50 kg * -2.8 m/s) + (1.3 kg * -1.5 m/s) + (m4 * v4)
Simplifying further:
0 = -1.4 kg·m/s - 1.95 kg·m/s + (m4 * v4)
Combining the terms:
0 = -3.35 kg·m/s + (m4 * v4)
Now, let's solve for the speed (v4) and direction of the third fragment:
Rearranging the equation:
3.35 kg·m/s = (m4 * v4)
Dividing both sides by m4:
v4 = 3.35 kg·m/s / m4
Since we don't have the specific value of m4, we cannot determine the exact speed (v4) and direction of the third fragment without additional information.
Whatever it takes to make the vector sum of the thee fragment moments equal to zero.
The mass of the third fragment is 1.2 kg.