Question

A 1 kg object is moving with a velocity of 6 m/s to the right. It collides and sticks to a 2 kg object moving with a velocity of 3 m/s in the same direction. How much kinetic energy is lost in the collision

Answer 1

Answer on Question #64004, Physics / Mechanics | Relativity

Question:

A 1 kg object is moving with a velocity of 6 m/s to the right. It collides and sticks to a 2 kg object moving with a velocity of 3 m/s in the same direction. How much kinetic energy is lost in the collision?

Solution:

According to the law of conservation of energy we may write that

$$E _ {k} ^ {1} + E _ {k} ^ {2} = E _ {k} ^ {1 2} + \Delta E$$ $$E _ {k} ^ {1} = \frac {m _ {1} v _ {1} ^ {2}}{2}, E _ {k} ^ {2} = \frac {m _ {2} v _ {2} ^ {2}}{2}, E _ {k} ^ {1 2} = \frac {(m _ {1} + m _ {2}) v _ {2} ^ {2}}{2}$$ $$\Delta E = E _ {k} ^ {1} + E _ {k} ^ {2} - E _ {k} ^ {1 2} = \frac {m _ {1} v _ {1} ^ {2} + m _ {2} v _ {2} ^ {2} - (m _ {1} + m _ {2}) v _ {2} ^ {2}}{2}$$ $$\Delta E = \frac {1 \cdot 6 ^ {2} + 2 \cdot 3 ^ {2} - (1 + 2) \cdot 3 ^ {2}}{2} = 13.5 \, \text{J}$$

Answer:

$$13.5 \, \text{J}$$

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