Two shuffleboard disks of equal mass, one orange and the other yellow, are invol

Question

Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in
a perfectly elastic, glancing collision. The yellow disk is initially at rest and is struck
by the orange disk moving with a speed of 3.64 m/s. After the collision, the orange disk
moves along a direction that makes an angle of 51.2 degrees with its initial direction of motion and the velocity of the yellow disk is perpendicular to that of the orange disk (after the collision).

Determine the final speed of the orange disk. Answer in units of m/s.

Explanation / Answer

Writing equations in component form from the law of conservation of momentum (calling the orange disc m1 and the yellow disc m2), the x-component equation is: P(i) = p(f) (m1v1)0 + (m2v2)0 = m1v1 + m2v2 m(7.04m/s) + 0 = mv1 cos? + mv2cos? 7.04m/s = v1cos33.3° + v2cos? v2cos? = 7.04m/s - v1cos33.3°-------------->eqn. (1) For the y-component: 0 = v1sin? + v2sin? v2sin? = -v1sin33.3°---------->eqn. (2) Recalling that sin? / cos? = tan?, divide (2) by (1): Tan? = -0.549v1 / (7.04m/s – 0.836v1)----------->eqn.(3) Now since the yellow disc’s velocity is perpendicular to the orange disc’s, the combined angle between them is 90.0°. So, 90.0° -33.3° = 56.7° (the angle of the yellow disc). But since the original line of motion of the orange disc lies on the x axis, this puts the yellow pucs angle of direction below the x axis, or -56.7°. Plug this into (3), and solve for v1: Tan(-56.7°) = -0.549v1 / (7.04m/s – 0.836v1) v1 = 5.88m/s.

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