Two disks of equal masses, one orange and the other yellow are involved in an elastic glancing collision. The yellow disk is initially at rest and is struck by an orange disk moving with a speed of 5 mathrm(~m) / mathrm(s) in the direction of the positive x -axis. After collision, the orongle disk moves along a direction that makes an angle of 37^circ with its initial direction of motion. The velocities of two disks are perpendicular after collision. If vec(v)_(0), vec(v)_(y), vec(V)_(0), vec(v)_(y) are initial and final velocities of the orunge and yellow disks respective and the total initial momentum is varepsilon vec(P)_(i)=5 mathrm(mi) , then determine the final speed of each disk.

To solve this problem, we need to use the principles of conservation of momentum and the properties of an elastic collision. Let's denote the initial velocity of the orange disk as \( \vec{U}_0 \) and the initial velocity of the yellow disk as \( \vec{V}_0 \). Since the yellow disk is initially at rest, \( \vec{V}_0 = 0 \).<br /><br />Given:<br />- Initial velocity of the orange disk: \( \vec{U}_0 = \hat{B} \, \text{m/s} \)<br />- Initial velocity of the yellow disk: \( \vec{V}_0 = 0 \)<br />- Angle between the final direction of the orange disk and its initial direction: \( \theta = 37^\circ \)<br />- Total initial momentum: \( \vec{P}_{\text{initial}} = 5 \, \text{kg} \cdot \text{m/s} \)<br /><br />Let's denote the final velocities of the orange and yellow disks as \( \vec{U} \) and \( \vec{V} \) respectively. Since the velocities are perpendicular after the collision, we can use the right-angle triangle formed by the initial and final velocities of the orange disk to find the final speed of the orange disk.<br /><br />Using the Pythagorean theorem:<br />\[ U_x^2 + U_y^2 = U^2 \]<br /><br />Where \( U_x \) is the component of the final velocity of the orange disk along the x-axis and \( U_y \) is the component along the y-axis. Since the orange disk makes an angle of \( 37^\circ \) with its initial direction, we have:<br />\[ U_x = U \cos(37^\circ) \]<br />\[ U_y = U \sin(37^\circ) \]<br /><br />Substituting these into the Pythagorean theorem:<br />\[ U^2 = (U \cos(37^\circ))^2 + (U \sin(37^\circ))^2 \]<br />\[ U^2 = U^2 (\cos^2(37^\circ) + \sin^2(37^\circ)) \]<br />\[ U^2 = U^2 (1) \]<br />\[ U = \sqrt{U^2} \]<br /><br />Since the total initial momentum is conserved, we have:<br />\[ \vec{P}_{\text{initial}} = \vec{P}_{\text{final}} \]<br />\[ \vec{U}_0 + \vec{V}_0 = \vec{U} + \vec{V} \]<br />\[ \hat{B} = \hat{U} + \hat{V} \]<br /><br />Since \( \vec{V}_0 = 0 \), we have:<br />\[ \hat{B} = \hat{U} + \hat{V} \]<br /><br />To find the final speed of the yellow disk, we use the conservation of momentum in the y-direction:<br />\[ P_{\text{initial, y}} = P_{\text{final, y}} \]<br />\[ 0 = U_y + V_y \]<br />\[ V_y = -U_y \]<br /><br />Since the magnitudes of the velocities are equal:<br />\[ V = U \]<br /><br />Therefore, the final speeds of the orange and yellow disks are:<br />\[ U = \sqrt{U^2} = \sqrt{(\hat{B})^2} = \hat{B} \, \text{m/s} \]<br />\[ V = U = \hat{B} \, \text{m/s} \]<br /><br />So, the final speed of each disk is \( \hat{B} \, \text{m/s} \).