Question

A 200-kg rocket in deep space moves with a velocity of (121m/s)i+(38.0m/s)j. Suddenly, it explodes into three pieces, with the first (78 kg) moving at −(321m/s)i+(228m/s)j and the second (56 kg) moving at (16.0m/s) i−(88.0m/s) j. Find the velocity of the third piece.

a) (224m/s) i−(178m/s) j
b) (−136m/s)i+(38m/s)j
c) (145m/s)i −(190m/s)j
d) (−145m/s)i+(190m/s)j

Answer 1

Using the principle of conservation of momentum and knowing the mass and velocity of the pieces of the rocket before and after explosion, the velocity of the third piece of the rocket is calculated to be (-(136 m/s) i + (38 m/s) j).

Step-by-step explanation:

To find the velocity of the third piece of the rocket after the explosion, we can apply the principle of conservation of momentum. Since there are no external forces acting on the system, the total momentum before the explosion must equal the total momentum after the explosion.

The initial momentum Pinitial of the rocket is given by its mass times its velocity:

Pinitial = (200 kg) *((121 m/s) i + (38 m/s) j)

The first piece's momentum P1 is:

P1 = (78 kg) * (-(321 m/s) i + (228 m/s) j)

The second piece's momentum P2 is:

P2 = (56 kg) * ((16.0 m/s) i - (88.0 m/s) j)

By conservation of momentum:

Pinitial = P1 + P2 + P3

Where P3 is the momentum of the third piece. We then solve for P3 and divide by the mass of the third piece (200 kg - 78 kg - 56 kg = 66 kg) to find its velocity.

After calculations, the velocity of the third piece is found to be option (b): (-(136 m/s) i + (38 m/s) j).