Question

Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in an elastic, glancing collision. The yellow disk is initially at rest and is struck by the orange disk moving with a speed of 5.60 m/s. After the collision, the orange disk moves along a direction that makes an angle of 36.0° with its initial direction of motion. The velocities of the two disks are perpendicular after the collision. Determine the final speed of each disk.

Answer 1

Using Momentum conservation, in vertical and horizontal direction:

In Horizontal direction:

Pix = Pfx

m1*u1x + m2*u2x = m1*v1x + m2*v2x

u1x = Speed of orange disk before collision = 5.60 m/sec

u2x = Speed of yellow disk before collision = 0 m/sec

v1x = Speed of orange disk after collision = v1*cos 36 deg

Since both disk are perpendicular after collision, So yellow disk will be at (90 - 36 = 54 deg) with the initial direction of motion below positive x-axis.

v2x = Speed of yellow disk after collision = v1*cos 54 deg

m1 = m2 = m

So,

u1x + u2x = v1x + v2x

5.6 + 0 = v1*cos 36 deg + v2*cos 54 deg

5.6 = v1*0.809 + v2*0.588 ............ eq 1

Now In vertical direction

Piy = Pfy

m1*u1y + m2*u2y = m1*v1y + m2*v2y

u1y = 0 m/sec & u2y = 0 m/sec

v1y = v1*sin 36 deg

v2y = -v2*sin 54 deg

m1 = m2

So,

0 + 0 = v1*sin 36 deg - v2*sin 54 deg

v1 = v2*(sin 54 deg)/(sin 36 deg) = 1.376*v2

v1 = 1.376*v2

Using above value in eq 1:

5.6 = v2*1.376*0.809 + v2*0.588

v2 = 5.6/(1.376*0.809 + 0.588)

v2 = 3.29 m/sec = final speed of yellow disk

v1 = 1.376*3.29

v1 = 4.53 m/sec = final speed of orange disk

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