Question

Two air track gliders experience an elastic head-on collision. A 0.28 kg glider is heading at 2.13 m/s when it collides with a 2.58 kg glider which is moving at -3.58 m/s.

What is the velocity of the 0.28 kg object after the collision?
What is the velocity of the 2.58 kg object after the collision?

Answer 1

Answer on Question #39104, Physics, Mechanics

Question:

Two air track gliders experience an elastic head-on collision. A 0.28 kg glider is heading at 2.13 m/s when it collides with a 2.58 kg glider which is moving at -3.58 m/s.

What is the velocity of the 0.28 kg object after the collision?

What is the velocity of the 2.58 kg object after the collision?

Answer:

The law of conservation of momentum:

$$m _ {1} v _ {1} + m _ {2} v _ {2} = m _ {1} v _ {1} ^ {\prime} + m _ {2} v _ {2} ^ {\prime}$$ $$m _ {1} (v _ {1} - v _ {1} ^ {\prime}) = m _ {2} (v _ {2} ^ {\prime} - v _ {2})$$

where $m_{1} = 0.28 \, kg, m_{2} = 2.58, v_{1}, v_{2}$ are speeds before collision, $v_{1}', v_{2}'$ - after.

The law of conservation of energy:

$$\frac {m _ {1} v _ {1} ^ {2}}{2} + \frac {m _ {2} v _ {2} ^ {2}}{2} = \frac {m _ {1} v _ {1} ^ {\prime 2}}{2} + \frac {m _ {2} v _ {2} ^ {\prime 2}}{2}$$ $$\frac {m _ {1} v _ {1} ^ {2}}{2} - \frac {m _ {1} v _ {1} ^ {\prime 2}}{2} = \frac {m _ {2} v _ {2} ^ {\prime 2}}{2} - \frac {m _ {2} v _ {2} ^ {2}}{2}$$ $$m _ {1} (v _ {1} - v _ {2}) (v _ {1} + v _ {2}) = m _ {2} (v _ {2} ^ {\prime} - v _ {2}) (v _ {2} ^ {\prime} + v _ {2})$$

So, we have system of equations:

$$\left\{ \begin{array}{c} m _ {1} (v _ {1} - v _ {1} ^ {\prime}) = m _ {2} (v _ {2} ^ {\prime} - v _ {2}) \\ m _ {1} (v _ {1} - v _ {1} ^ {\prime}) (v _ {1} + v _ {1} ^ {\prime}) = m _ {2} (v _ {2} ^ {\prime} - v _ {2}) (v _ {2} ^ {\prime} + v _ {2}) \end{array} \right.$$

Or:

$$\left\{ \begin{array}{c} m _ {1} (v _ {1} - v _ {1} ^ {\prime}) = m _ {2} (v _ {2} ^ {\prime} - v _ {2}) \\ (v _ {1} + v _ {1} ^ {\prime}) = (v _ {2} ^ {\prime} + v _ {2}) \end{array} \right.$$

Solving for $v_{1}^{\prime}$ and $v_{2}^{\prime}$:

$$v _ {1} ^ {\prime} = \frac {2 m _ {2} v _ {2} + (m _ {1} - m _ {2}) v _ {1}}{m _ {1} + m _ {2}} = - 8. 1 7 \frac {m}{s}$$ $$v _ {2} ^ {\prime} = \frac {2 m _ {1} v _ {1} - (m _ {1} - m _ {2}) v _ {2}}{m _ {1} + m _ {2}} = - 2.46 \frac {m}{s}$$

Answer: $v_{1}^{\prime} = -8.17\frac{m}{s}, v_{2}^{\prime} = -2.46\frac{m}{s}$