Answer :
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].
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].
