An object at rest explodes into three pieces of equal mass

An object at rest explodes into three pieces of equal mass. The first piece moves east at 30 m/s, and the second northwest at 10 m/s. Find the modulus and orientation of the velocity of the third piece.

The correct answer and explanation is:

Let’s analyze the problem step-by-step:

Problem summary:

• An object initially at rest breaks into 3 pieces of equal mass.
• Piece 1 moves east at 30 m/s.
• Piece 2 moves northwest at 10 m/s.
• Find the velocity (magnitude and direction) of the third piece.

Step 1: Understand the physics principle

Since the object was initially at rest, its total initial momentum was zero.

By conservation of momentum, the vector sum of the momenta of the three pieces after the explosion must also be zero: p⃗1+p⃗2+p⃗3=0\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0

Each piece has equal mass mm, so momentum p⃗=mv⃗\vec{p} = m \vec{v}.

Dividing the equation by mm (same for all), we get: v⃗1+v⃗2+v⃗3=0\vec{v}_1 + \vec{v}_2 + \vec{v}_3 = 0

So, v⃗3=−(v⃗1+v⃗2)\vec{v}_3 = – (\vec{v}_1 + \vec{v}_2)

Step 2: Express the given velocities as vectors

• Define the positive x-axis pointing east, positive y-axis pointing north.

1. First piece (east at 30 m/s):

v⃗1=(30,0)m/s\vec{v}_1 = (30, 0) \quad \text{m/s}

2. Second piece (northwest at 10 m/s):

Northwest means 45° between north and west.

So, components: v2x=10cos⁡135∘=10×(−22)=−7.07 m/sv_{2x} = 10 \cos 135^\circ = 10 \times (-\frac{\sqrt{2}}{2}) = -7.07 \, \text{m/s} v2y=10sin⁡135∘=10×22=7.07 m/sv_{2y} = 10 \sin 135^\circ = 10 \times \frac{\sqrt{2}}{2} = 7.07 \, \text{m/s}

So, v⃗2=(−7.07,7.07)m/s\vec{v}_2 = (-7.07, 7.07) \quad \text{m/s}

Step 3: Calculate v⃗3\vec{v}_3

v⃗3=−(v⃗1+v⃗2)=−((30,0)+(−7.07,7.07))=−(22.93,7.07)=(−22.93,−7.07) m/s\vec{v}_3 = – (\vec{v}_1 + \vec{v}_2) = – \big( (30, 0) + (-7.07, 7.07) \big) = – (22.93, 7.07) = (-22.93, -7.07) \, \text{m/s}

Step 4: Calculate magnitude and direction of v⃗3\vec{v}_3

Magnitude: ∣v⃗3∣=(−22.93)2+(−7.07)2=525.8+50=575.8≈24.0 m/s|\vec{v}_3| = \sqrt{(-22.93)^2 + (-7.07)^2} = \sqrt{525.8 + 50} = \sqrt{575.8} \approx 24.0 \, \text{m/s}

Direction:

Find the angle θ\theta with respect to the positive x-axis (east), measured counterclockwise: θ=tan⁡−1(v3yv3x)=tan⁡−1(−7.07−22.93)=tan⁡−1(0.308)=17.1∘\theta = \tan^{-1} \left(\frac{v_{3y}}{v_{3x}}\right) = \tan^{-1} \left(\frac{-7.07}{-22.93}\right) = \tan^{-1}(0.308) = 17.1^\circ

Since both components are negative, v⃗3\vec{v}_3 lies in the third quadrant.

So, angle from the positive x-axis is: θ=180∘+17.1∘=197.1∘\theta = 180^\circ + 17.1^\circ = 197.1^\circ

Alternatively, you can say the velocity is 17.1° south of west.

Final answer:

• Velocity magnitude of the third piece = 24.0 m/s
• Direction = 197.1° from east (or 17.1° south of west).

Explanation (about 300 words):

This problem illustrates the principle of conservation of momentum in two dimensions. Since the original object was at rest, its initial momentum was zero. When the object explodes into three pieces, the vector sum of the momenta of all pieces must remain zero to satisfy conservation laws.

Given that the masses are equal, we focus on the velocity vectors themselves. The first piece moves due east at 30 m/s, and the second piece moves northwest (which is 135° from the positive x-axis) at 10 m/s. By resolving the second piece’s velocity into x and y components, we obtain its vector form as (-7.07 m/s, 7.07 m/s).

The third piece’s velocity vector must be such that when added to the first two, the total vector sum equals zero. Hence, the third piece’s velocity is the negative of the sum of the first two velocity vectors. Calculating this results in (-22.93 m/s, -7.07 m/s).

To find the magnitude of the third piece’s velocity, we use the Pythagorean theorem on its components. The direction is found using the arctangent of the y-component over the x-component, considering the quadrant of the vector.

The final velocity of the third piece is approximately 24 m/s, heading 17.1° south of west. This direction and magnitude balance the momentum such that the total momentum after the explosion remains zero, consistent with the law of conservation of momentum.