Question

A 27.6-g object moving to the right with a speed of 23.9 cm/s overtakes and collides elastically with a 18.1-g object moving in the same direction with a speed of 14.9 cm/s. Find the velocity (in cm/s) of 18.1-g object after the collision.

Answer 1

Answer on Question#39105 – Physics – Mechanics

A 27.6-g object moving to the right with a speed of $23.9~\mathrm{cm/s}$ overtakes and collides elastically with a 18.1-g object moving in the same direction with a speed of 14.9 cm/s. Find the velocity (in cm/s) of 18.1-g object after the collision.

Solution:

$m_{1} = 27.6g$ is mass of the first object;

$v_{1} = 23.9\frac{cm}{s}$ is the initial velocity of $27.6g$ object

$v_{1}^{\prime}$ is final velocity of $27.6g$ object

$m_{2} = 18.1g$ is mass of the second object;

$v_{2} = 14.9\frac{cm}{s}$ is the initial velocity of $18.1g$ object

$v_{2}^{\prime}$ is final velocity of $18.1g$ object

This is a conservation of momentum and energy problem.

Conservation of momentum before and after collision:

$$x: m_{1}v_{1} + m_{2}v_{2} = m_{1}v_{1}^{\prime} + m_{2}v_{2}^{\prime}$$ $$m_{1}(v_{1} - v_{1}^{\prime}) = m_{2}(v_{2}^{\prime} - v_{2})$$

Conservation of kinetic energy before and after collision:

$$\frac{m_{1}v_{1}^{2}}{2} + \frac{m_{2}v_{2}^{2}}{2} = \frac{m_{1}v_{1}^{\prime 2}}{2} + \frac{m_{2}v_{2}^{\prime 2}}{2}$$ $$m_{1}\left(v_{1}^{2} - v_{1}^{\prime 2}\right) = m_{2}\left(v_{2}^{\prime 2} - v_{2}^{2}\right)$$

Using the difference of squares formula:

$$m_{1}(v_{1} - v_{1}^{\prime})(v_{1} + v_{1}^{\prime}) = m_{2}(v_{2}^{\prime} - v_{2})(v_{2}^{\prime} + v_{2})$$ $$(2) \div (1)$$ $$\frac{m_{1}(v_{1} - v_{1}^{\prime})(v_{1} + v_{1}^{\prime})}{m_{1}(v_{1} - v_{1}^{\prime})} = \frac{m_{2}(v_{2}^{\prime} - v_{2})(v_{2}^{\prime} + v_{2})}{m_{2}(v_{2}^{\prime} - v_{2})}$$ $$v_{1} + v_{1}^{\prime} = v_{2}^{\prime} + v_{2}$$ $$v_{1}^{\prime} = v_{2}^{\prime} + v_{2} - v_{1}$$ $$(3)\text{in}(1):$$ $$m_{1}(v_{1} - (v_{2}^{\prime} + v_{2} - v_{1})) = m_{2}(v_{2}^{\prime} - v_{2})$$ $$m_{1}v_{1} - m_{1}v_{2}^{\prime} - m_{1}v_{2} + m_{1}v_{1} = m_{2}v_{2}^{\prime} - m_{2}v_{2}$$ $$v_{2}^{\prime}(m_{1} + m_{2}) = v_{2}(m_{1} - m_{2}) - 2m_{1}v_{1}$$ $$v_{2}^{\prime} = \frac{v_{2}(m_{1} - m_{2}) - 2m_{1}v_{1}}{m_{1} + m_{2}} =$$ $$= \frac{14.9 \frac{cm}{s} (27.6g - 18.1g) - 2 \cdot 27.6g \cdot 23.9 \frac{cm}{s}}{27.6g + 18.1g} = -25.8 \frac{cm}{s}$$

Answer: the velocity of 18.1-g object after the collision is equal to $25.8\frac{cm}{s}$.