Question

A particle mass m, traveling at a speed of v, strikes a stationary particle mass 2m. As a result the partucle of mass m, us deflected through an angle of 45 degrees at a speed of v/2. What is the speed and direction of mass 2m?

Answer 1

Answer on Question #85415 - Physics - Molecular Physics | Thermodynamics

A particle mass $m$ , traveling at a speed of $v$ , strikes a stationary particle mass $2m$ . As a result the particle of mass $m$ , is deflected through an angle of 45 degrees at a speed of $v/2$ . What is the speed and direction of mass $2m$ ?

Solution.

According to the law of conservation of momentum, the total momentum of an isolated system is constant:

$$m \vec {\vartheta} + 0 = \frac {1}{2} m \vec {\vartheta} + 2 m \vec {\vartheta} ^ {\prime}$$ $$\vec {\vartheta} = \frac {1}{2} \vec {\vartheta} + 2 \vec {\vartheta} ^ {\prime}$$

Let's find projections of (1) into coordinate axes:

Ox axis:

$$\vartheta = \frac {1}{2} \vartheta \cos 4 5 {}^ {\circ} + 2 \vartheta^ {\prime} \cos \theta$$

Oy axis:

$$\frac {1}{2} \vartheta \sin 4 5 {}^ {\circ} = 2 \vartheta^ {\prime} \sin \theta$$

From (3):

$$\vartheta^ {\prime} = \frac {\vartheta \sin 4 5 {}^ {\circ}}{4 \sin \theta}$$

Let's substitute (4) into (2):

$$\vartheta = \frac {1}{2} \vartheta \cos 4 5 {}^ {\circ} + 2 \frac {\vartheta \sin 4 5 {}^ {\circ}}{4 \sin \theta} \cos \theta$$ $$2 = \cos 4 5 {}^ {\circ} + \frac {\sin 4 5 {}^ {\circ}}{\tan \theta}$$ $$\frac {\sin 4 5 {}^ {\circ}}{\tan \theta} = 2 - \cos 4 5 {}^ {\circ}$$ $$\tan \theta = \frac {\sin 4 5 {}^ {\circ}}{2 - \cos 4 5 {}^ {\circ}} = \frac {0 . 7 0 7 1}{1 . 2 9 2 9} = 0. 5 4 6 9$$ $$\theta = 2 9 {}^ {\circ}$$

Let's substitute the value of $\theta$ into (4):

$$\vartheta^ {\prime} = \frac {\vartheta \sin 4 5 {}^ {\circ}}{4 \sin 2 9 {}^ {\circ}} = \frac {0 . 7 0 7 1}{1 . 9 3 9 2} \vartheta = 0. 4 \vartheta$$

Answer: $\vartheta' = 0.4\vartheta, \theta = 29{}^\circ$

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