Question

two shuffleboard of equal mass,one orange and the other yellow,are involved in a elastic,glancing collision.the yellow disk is initially at rest and is struck by the orange disk moving with a speed of 5.00 m/s.after collision,the orange disk moves along a direction that makes an angle of 37.0 with its initial direction of motion.the velocities of the two disks are perpendicular after the collusion.determine the final speed of each disk

Answer 1

Solution.

Before collision:

$mv1 + mv2 = const$

v2 = 0, so

$mv1 = const$

After collision:

Horizontal direction

$mv1 = mv^{'}1cos(37^o)+mv^{'}2sin(90^o-53^o)$

$v1 = v^{'}1cos(37^o)+v^{'}2sin(90^o-53^o)$

Vertical direction

$v^{'}1sin(37^o)-v^{'}2sin(53^o) = 0$

$v^{'}1 = v^{'}2 \frac{sin(53^o)}{sin(37^o)}$

$v1 = v^{'}2 \frac{sin(53^o)}{sin(37^o)}cos(37^o)+v^{'}2cos(53^o)$

$v1sin(37^o) = v^{'}2$

$v2' = 5 \times sin(37^o) = 3.01 m/s$

$v^{'}1 = 3.01 \frac{sin(53^o)}{sin(37^o)} = 3.99 m/s$

Answer:

v(orange) = 3.99 m/s

v(yellow) = 3.01 m/s