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Two pieces of clay collide and stick together, moving in one dimension. Clay 1 has a mass of 2,100 grams, and clay 2 has a mass of 2,500 grams. Clay 1 has an initial velocity of [tex]$20 \frac{m}{s}$[/tex] and clay 2 has an initial velocity of [tex]$-10 \frac{m}{s}$[/tex] immediately before they have a perfectly inelastic collision.

What is the final velocity of the two pieces of clay immediately after the collision?

Calculate the final velocity.


Sagot :

To find the final velocity of the two pieces of clay immediately after a perfectly inelastic collision, we need to use the principle of conservation of momentum. In a perfectly inelastic collision, the two objects stick together and move with the same final velocity after the collision.

Given:
- Mass of clay 1 ([tex]\( m_1 \)[/tex]) = 2,100 grams = 2.1 kg (since 1 kg = 1,000 grams)
- Mass of clay 2 ([tex]\( m_2 \)[/tex]) = 2,500 grams = 2.5 kg
- Initial velocity of clay 1 ([tex]\( v_1 \)[/tex]) = 20 m/s
- Initial velocity of clay 2 ([tex]\( v_2 \)[/tex]) = -10 m/s

The formula for the final velocity ([tex]\( v_f \)[/tex]) in a perfectly inelastic collision is:

[tex]\[ v_f = \frac{m_1 \cdot v_1 + m_2 \cdot v_2}{m_1 + m_2} \][/tex]

Substituting the given values:

[tex]\[ v_f = \frac{(2.1 \, \text{kg} \cdot 20 \, \text{m/s}) + (2.5 \, \text{kg} \cdot (-10) \, \text{m/s})}{2.1 \, \text{kg} + 2.5 \, \text{kg}} \][/tex]

Calculate the numerator and the denominator separately:

Numerator:

[tex]\[ (2.1 \, \text{kg} \cdot 20 \, \text{m/s}) + (2.5 \, \text{kg} \cdot (-10) \, \text{m/s}) = 42 \, \text{kg} \cdot \text{m/s} + (-25 \, \text{kg} \cdot \text{m/s}) = 42 \, \text{kg} \cdot \text{m/s} - 25 \, \text{kg} \cdot \text{m/s} = 17 \, \text{kg} \cdot \text{m/s} \][/tex]

Denominator:

[tex]\[ 2.1 \, \text{kg} + 2.5 \, \text{kg} = 4.6 \, \text{kg} \][/tex]

Now, calculate the final velocity:

[tex]\[ v_f = \frac{17 \, \text{kg} \cdot \text{m/s}}{4.6 \, \text{kg}} \approx 3.70 \, \text{m/s} \][/tex]

Therefore, the final velocity of the two pieces of clay immediately after the collision is approximately [tex]\( 3.70 \frac{m}{s} \)[/tex].