Question

a small explosive explodes into three pieces, they fly 120 degrees relative to one another.

the 1st piece has a mass of 6 kg.

Pieces 2 has a mass of 7 kg, and flies with a speed of 3 m/s.

Piece 3 has a mass of 7 kg, and flies with a speed of 3 m/s.

how fast does the 1st piece go in m/s?

Please explain how to solve (dumb it down) so that I could learn. Thank you!

Answer 1

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Answer 2

Given:

• An explosive initially at rest explodes into three pieces.

• The pieces fly apart at 120° relative to each other.

• Masses and velocities:

• Piece 1: $m_1=6\text{kg}$, $v_1=?$ (to find)

• Piece 2: $m_2=7\text{kg}$, $v_2=3\text{m/s}$

• Piece 3: $m_3=7\text{kg}$, $v_3=3\text{m/s}$

Key Physics Concept: Conservation of Momentum

Since the explosive was initially at rest, the total momentum before the explosion is zero. After the explosion, the vector sum of the momenta of all three pieces must also be zero:

$${\vec{p}}_1+{\vec{p}}_2+{\vec{p}}_3=0$$

Step 1: Assign Directions for Simplification

To handle the 120° angles, we set up a coordinate system:

• Let Piece 1 move along the +x-axis.

• Piece 2 moves at 120° from Piece 1.

• Piece 3 moves at 240° (or -120°) from Piece 1.

This ensures symmetry in the problem.

Step 2: Compute Momenta of Each Piece

Momentum ($\vec{p}$) is mass ($m$) times velocity ($\vec{v}$):

$$\vec{p}=m\vec{v}$$

1. Piece 1 (along +x-axis):

   $${\vec{p}}_1=6v_1\widehat{i}$$

   (No y-component since it's purely along the x-axis.)

2. Piece 2 (120° from +x-axis):

   $${\vec{p}}_2=7\times 3=21$$

• x-component: $p_{2x}=21\cos (120\mathrm{°})$

• y-component: $p_{2y}=21\sin (120\mathrm{°})$

3. Piece 3 (240° from +x-axis):

   $${\vec{p}}_3=7\times 3=21$$

• x-component: $p_{3x}=21\cos (240\mathrm{°})$

• y-component: $p_{3y}=21\sin (240\mathrm{°})$

Step 3: Calculate Trigonometric Components

Recall:

$$\cos (120\mathrm{°})=\cos (240\mathrm{°})=-\frac{1}{2}$$$$\sin (120\mathrm{°})=\frac{\sqrt{3}}{2},\sin (240\mathrm{°})=-\frac{\sqrt{3}}{2}$$

Thus:

• Piece 2:

  $$p_{2x}=21\times (-\frac{1}{2})=-10.5$$$$p_{2y}=21\times \frac{\sqrt{3}}{2}\approx 18.186$$

• Piece 3:

  $$p_{3x}=21\times (-\frac{1}{2})=-10.5$$$$p_{3y}=21\times (-\frac{\sqrt{3}}{2})\approx -18.186$$

Step 4: Apply Momentum Conservation in x and y Directions

1. In the x-direction:

   $$p_{1x}+p_{2x}+p_{3x}=0$$$$6v_1+(-10.5)+(-10.5)=0$$$$6v_1-21=0$$$$6v_1=21$$$$v_1=\frac{21}{6}=3.5\text{m/s}$$

2. In the y-direction:

   $$p_{1y}+p_{2y}+p_{3y}=0$$$$0+18.186+(-18.186)=0$$

   (This confirms that the y-components cancel out, as expected.)

3. 
   

Conclusion:

The first piece must move at 3.5 m/s along the x-axis to conserve momentum.