A spaceship of mass 2.1×106 kg is cruising at a speed of 5.8×106 m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 4.6×105 kg, is blown straight backward with a speed of 2.5×106 m/s. A second piece, with mass 7.9×105 kg, continues forward at 1.3×106 m/s.

Part A
What is the speed of the third piece?
Express your answer using two significant figures. m/s

Part B
What is the directionof the third piece?

Straight backward

Explanation:

Given a space ship of mass

M = 2.1 × 10^6 kg

The space ship is moving at

V = 5.8 × 10^6 m/s

This space ship blows into three

The first part is blown backward I.e. in the negative direction

Then, mass of first part and it's velocity is

M1 = 4.6 × 10^5 kg

V1 = -2.5 × 10^6 m/s

The second part continues in forward motion, I.e positive direction

Then, mass of the second part and it's velocity is

M2 = 7.9 × 10^5 kg

V2 = 1.3 × 10^6 m/s

A. What is the speed of the third piece?

So, we need to know the mass of the third part

M1 + M2 + M3 = M

M3 = M-M1-M2

M3 = 2.1 × 10^6 - 4.6 × 10^5 - 7.9 × 10^5

M3 = 8.5 × 10^5 kg

So, applying conservation of momentum

Momentum before explosion = momentum after explosion

MV = M1•V1 + M2•V2 + M3•V3

2.1 × 10^6 × 5.8 × 10^6 = 4.6 × 10^5 × -2.5 × 10^6 + 7.9 × 10^5 × 1.3 × 10^6 + 8.5×10^5•V3

1.218 × 10¹³ = -1.15 × 10¹² + 1.027 × 10¹² + 8.5 × 10^5•V3

1.218 × 10¹³ + 1.15 × 10¹² - 1.027 × 10¹² = 8.5 × 10^5•V3

1.2303 × 10^13 = 8.5 × 10^5 V3

V3 = 1.2303 × 10¹³ / 8.5 × 10^5

V3 = 1.45 × 10^7 m/s

B. The direction is straight forward since the velocity of the part is positive

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