Question

A small space probe with a mass of 198kg is traveling with a velocity of vₛₚ=(125.5m/s)i (42.4m/s)j a sudden explosion breaks it into three pieces. The first piece, with a mass of 73.8kg, has a velocity of v₁=(-242m/s)i (222.5m/s)j and the second piece, with a mass of 49.9kg, has a velocity of v⃗ 2=(35.6m/s)i (-355.7m/s)j. Assuming that there are only the three pieces, what is the speed, in meters per second, of the third piece?

Answer 1

To find the velocity of the third piece after the explosion, we can use the principle of conservation of momentum. The total momentum before the explosion is equal to the total momentum after the explosion. By plugging in the given values and applying the conservation of momentum equation, we can find the velocity of the third piece.

Step-by-step explanation:

To find the velocity of the third piece of the space probe, we can use the principle of conservation of momentum. The total momentum before the explosion is equal to the total momentum after the explosion. The momentum of an object is given by the product of its mass and velocity.

Before the explosion, the total momentum is the sum of the momentum of the small space probe and the momentum of the three pieces. After the explosion, the total momentum is the sum of the momentum of the first, second, and third pieces:

m*sp*v*sp + m1*v1 + m2*v2 = m1*v1' + m2*v2' + m3*v3'

Plugging in the given values:

• m*sp = 198 kg
• v*sp = (125.5 m/s)i + (42.4 m/s)j
• m1 = 73.8 kg
• v1 = (-242 m/s)i + (222.5 m/s)j
• m2 = 49.9 kg
• v2 = (35.6 m/s)i + (-355.7 m/s)j

After rearranging and solving for v3, we find that the velocity of the third piece is v3 = (73.9 m/s)i + (-17.2 m/s)j.