Question

An elastic collision occurs in one dimension, in which a 10kg block traveling at 5m/s collides with a 5kg block traveling at 3m/s in the same direction. What are the velocities of the two blocks immediately after the collision?

Answer 1

Answer on Question #77086, Physics, Mechanics, Relativity

An elastic collision occurs in one dimension, in which a 10 kg block traveling at 5 m/s collides with a 5 kg block traveling at 3 m/s in the same direction.

What are the velocities of the two blocks immediately after the collision?

Solution

From the law of conservation of momentum

$$m _ {1} \cdot v _ {1 i} + m _ {2} \cdot v _ {2 i} = m _ {1} \cdot v _ {1} + m _ {2} \cdot v _ {2}$$

where $v_{1i}$ - velocities before collision are $v_{1i} = 5 \, \text{m/s}$. Particles of $m_1 = 10 \, \text{kg}$

$v_{2i}$ - velocities before collision are $v_{2i} = 3 \, \text{m/s}$. Particles of $m_2 = 5 \, \text{kg}$

$v_{1}$ - velocities after collision. Particles of $m_1 = 10 \, \text{kg}$

$v_{2}$ - velocities after collision. Particles of $m_2 = 5 \, \text{kg}$

Energy conservation law

$$\frac {m _ {1} \cdot v _ {1 i} ^ {2}}{2} + \frac {m _ {2} \cdot v _ {2 i} ^ {2}}{2} = \frac {m _ {1} \cdot v _ {1} ^ {2}}{2} + \frac {m _ {2} \cdot v _ {2} ^ {2}}{2}$$

From here

$$v _ {1} = \frac {v _ {1 i} \cdot (m _ {1} - m _ {2})}{m _ {1} + m _ {2}} + \frac {v _ {2 i} \cdot 2 \cdot m _ {2}}{m _ {1} + m _ {2}} = \frac {5 \cdot (10 - 5)}{10 + 5} + \frac {3 \cdot 2 \cdot 5}{10 + 5} = 3.66 \, \frac {m}{s}$$ $$v _ {2} = \frac {v _ {1 i} \cdot 2 \cdot m _ {1}}{m _ {1} + m _ {2}} + \frac {v _ {2 i} \cdot (m _ {2} - m _ {1})}{m _ {1} + m _ {2}} = \frac {5 \cdot 2 \cdot 10}{10 + 5} + \frac {3 \cdot (5 - 10)}{10 + 5} = 5.66 \, \frac {m}{s}$$

Answer: $v_{1} = 3.66 \, \frac{m}{s}, \, v_{2} = 5.66 \, \frac{m}{s}$

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