Two shuffleboard disks of equal mass, one orange and the other green, are involv

Question

Two shuffleboard disks of equal mass, one orange and the other green, are involved in a perfectly elastic glancing collision. The green disk is initially at rest and is struck by the orange disk moving initially to the right at vOi = 6.10 m/s as in Figure a, shown below. After the collision, the orange disk moves in a direction that makes an angle of = 35.0° with the horizontal axis while the green disk makes an angle of = 55.0° with this axis as in Figure b. Determine the speed of each disk after the collision.

vof = m/s

vgf = m/s

Explanation / Answer

Suppose the orange disc is moving in the +ve x-direction and after the collision gets deflected to 35° above x-axis and the yellow disc gets deflected to 90° - 35° = 55° below the x-axis.
Let u = velocity of the orange disc after the collision
and v = velocity of the yellow disc after the collision.

By the law of conservation of linear momentum applied along the +ve x-axis
m * 5.90 + 0 = m * u * cos35° + m * vcos55°
=> ucos35° + vcos53.6° = 5.90
=> u (cos35°)/(cos55°) = (5.90)/(cos53.6°)
=> 1.3564 u + v = 9.9424 ... ( 1 )

By the law of conservation of linear momentum applied along the +ve y-axis
m * usin35° = m * vsin55°
=> u (sin35°)/(sin55°) - v = 0
=> 0.7373u - v = 0 ... ( 2 )

Adding eqns. ( 1 ) and ( 2 ),
2.0937u = 9.9424
=> u = 4.7487 m/s

From eqn. ( 2 )
v = (0.7373) * (4.7487) = 3.5012 m/s

======================================...

In the case of elastic collision, they will both move perpendicular to each is the property of this collision. To check the above answer, we can verify by checking if the law of conservation of KE is satisfied.
Initial KE = (1/2) m * (5.9)^2 = 17.405 m joule (m = mass of the disc)
Final KE = (1/2) m * (4.7487)^2 + (1/2) m (3.5012)^2 = 17.404 joule

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