two shuffleboard disks of equal mass, one orange and the other yellow are involv
ID: 1364398 • Letter: T
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 initially to the right at 5.90 m/s. After the collision, the orange disk moves in a direction that makes an angle of 34.6 degrees with its initial direction. Meanwhile, the velocity vector of the yellow disk is perpendicular to the postcollision velocity vector of the orange disk. Determine the speed of each disk after the collision. 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 initially to the right at 5.90 m/s. After the collision, the orange disk moves in a direction that makes an angle of 34.6 degrees with its initial direction. Meanwhile, the velocity vector of the yellow disk is perpendicular to the postcollision velocity vector of the orange disk. Determine the speed of each disk after the collision. 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 initially to the right at 5.90 m/s. After the collision, the orange disk moves in a direction that makes an angle of 34.6 degrees with its initial direction. Meanwhile, the velocity vector of the yellow disk is perpendicular to the postcollision velocity vector of the orange disk. Determine the speed of each disk after the collision.Explanation / Answer
Suppose the orange disc is moving in the +ve x-direction and after the collision gets deflected to 34.6 above x-axis and the yellow disc gets deflected to 90 - 34.6 = 55.4 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 * cos34.6 + m * vcos55.4
=> ucos34.6 + vcos55.4 = 5.90
=> u (cos34.6)/(cos55.4) = (5.90)/(cos55.4)
=> 1.45 u + v = 10.39 ... ( 1 )
By the law of conservation of linear momentum applied along the +ve y-axis
m * usin34.6 = m * vsin55.4
=> u (sin34.6)/(sin55.4) - v = 0
=> 0.69u - v = 0 ... ( 2 )
Adding eqns. ( 1 ) and ( 2 ),
2.14u = 10.39
=> u = 4.85 m/s
From eqn. ( 2 )
v = (0.69) * (4.85) = 3.35 m/s
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.