Question

6. a) The distance between the oxygen molecule and each of the hydrogen atoms in a water (H2O) molecule is 0.96 Å and the angle between the two oxygen-hydrogen bonds is 105°. Treating the atoms as particles, find the centre of mass of the system.


b) A stationary ball, with a mass of 0.2 kg, is struck by an identical ball moving at .4ms −1 After the collision, the second ball moves 60° to the left of its original direction. The stationary ball moves 30° to the right of the moving ball’s original direction. What is the velocity of each ball after the collision?

Answer 1

Answer on Question #40138, Physics, Mechanics | Kinematics | Dynamics

a) The distance between the oxygen molecule and each of the hydrogen atoms in a water (H2O) molecule is 0.96 Å and the angle between the two oxygen-hydrogen bonds is 105°. Treating the atoms as particles, find the centre of mass of the system.

b) A stationary ball, with a mass of 0.2 kg, is struck by an identical ball moving at .4 ms⁻¹. After the collision, the second ball moves 60° to the left of its original direction. The stationary ball moves 30° to the right of the moving ball's original direction. What is the velocity of each ball after the collision?

Solution:

a)

The first step is to choose a coordinate system, such as the one in the diagram, and locate each particle. The chosen origin is the centre of the box.

The coordinates of the centre of mass are given by

$$x_{cm} = \frac{\sum m_{i} x_{i}}{M_{total}} = \frac{0.711534}{18} = 0.0395297 = 0.0395\ \text{\AA}$$

And

$$y_{cm} = \frac{\sum m_{i} y_{i}}{M_{total}} = \frac{0.927289}{18} = 0.051516 = 0.0515\ \text{\AA}$$

Answers will vary based on the choice of coordinate system.

b) Solution:

$u = 4.0 \, \text{m/s}$ is initial speed of second ball

Let the velocity of the stationary ball and moving ball is $v_1$ and $v_2$ m/s respectively after the collision

By the law of momentum conservation in X-Y plane:

$$m v_1 \cos 30{}^\circ + m v_2 \cos 60{}^\circ = m u$$ $$v_1 \cos 30{}^\circ + v_2 \cos 60{}^\circ = u$$ $$m v_1 \sin 30{}^\circ = m v_2 \sin 60{}^\circ$$

From (2):

$$0.5 v_1 = 0.866 v_2$$

$v_1 = 1.732 v_2$ by putting this value in (1):

$$1.732 v_2 \times 0.866 + v_2 \times 0.5 = 4.0$$ $$v_2 = 4.0 / (1.732 \times 0.866 + 0.5) = 2.0$$ $$v_2 = 2.0 \, \text{m/s}$$

Hence $v_1 = 3.464 \, \text{m/s}$

Answer. a) The centre of mass is located at $(0.0395 \, \text{\AA}, 0.0515 \, \text{\AA})$.

b) final speed of initially stationary ball $3.46 \, \text{m/s}$,

final speed of initially moving ball $2.0 \, \text{m/s}$