Question

Perfectly elastic oblique collision occurs between a ball A moving along the x-axis and a ball B at rest and of the same mass as ball A. After the collision, ball A moves at an angle of 30° with the X direction and ball B at an angle theta with the X axis. The value of theta is?

Answer 1

Answer on Question #76289-Physics-Mechanics-Relativity

Perfectly elastic oblique collision occurs between a ball A moving along the x-axis and a ball B at rest and of the same mass as ball A. After the collision, ball A moves at an angle of $30{}^{\circ}$ with the X direction and ball B at an angle theta with the X axis. The value of theta is?

Solution

From the conservation of momentum:

X-direction:

$$mv = mv_a \cos 30 + mv_b \cos \theta$$ $$v = v_a \cos 30 + v_b \cos \theta$$

Y-direction:

$$0 = mv_a \sin 30 - mv_b \sin \theta$$ $$v_a \sin 30 = v_b \sin \theta$$

From the conservation of energy:

$$\frac{m v^2}{2} = \frac{m v_a^2}{2} + \frac{m v_b^2}{2}$$ $$v^2 = v_a^2 + v_b^2$$

Thus,

$$v = v_a \cos 30 + v_a \sin 30 \cot \theta = v_a (\cos 30 + \sin 30 \cot \theta)$$ $$v^2 = \left(\frac{v}{(\cos 30 + \sin 30 \cot \theta)}\right)^2 + \left(\frac{v \sin 30}{(\cos 30 + \sin 30 \cot \theta) \sin \theta}\right)^2$$ $$1 = \left(\frac{1}{(\cos 30 + \sin 30 \cot \theta)}\right)^2 + \left(\frac{\sin 30}{(\cos 30 + \sin 30 \cot \theta) \sin \theta}\right)^2$$ $$1 = \frac{1}{(\cos 30 + \sin 30 \cot \theta)^2} \left(1 + \frac{\sin^2 30}{\sin^2 \theta}\right)$$ $$1 = \frac{1}{(\cos 30 \sin \theta + \sin 30 \cos \theta)^2} (\sin^2 \theta + \sin^2 30)$$ $$1 = \frac{1}{(\sin (\theta + 30))^2} (\sin^2 \theta + \sin^2 30)$$

Thus,

$$\theta = 60{}^\circ$$

Answer: 60°.

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