A small ball of mass 0.1 kg is attached to a ℓ = 1 m-long thread and is hung to a horizontal peg. The small ball is at rest and another small ball of mass 0.2 kg, is projected from the ground and collides with it, such that the collision is totally elastic and head on, and after the collision the ball at the end of the thread oscillates with an angular amplitude of 90°. The distance between the peg and the ground is 2ℓ. What is distance x (in cm) shown in the figure?

Engineering

Physics

SHM of Simple Pendulum

Collision

Projectile Motion

Question

A small ball of mass 0.1 kg is attached to a ℓ = 1 m-long thread and is hung to a horizontal peg. The small ball is at rest and another small ball of mass 0.2 kg, is projected from the ground and collides with it, such that the collision is totally elastic and head on, and after the collision the ball at the end of the thread oscillates with an angular amplitude of 90°. The distance between the peg and the ground is 2ℓ.
What is distance x (in cm) shown in the figure?

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$1 = ℓ = \frac{v_{0}^{2} \sin^{2} α}{2 g}$

$e = 1 = \frac{\sqrt{20} - v}{v_{0} \cos α}$

$0 .2 V_{a} \cos α = 0 .1 \sqrt{20} + 0 .2 (\sqrt{20} - v_{0} \cos α)$

$0 .4 v_{0} \cos α = 0 .3 \sqrt{20}$

$V_{0} \cos α = \frac{3}{4} \sqrt{20}$

$x = \frac{v_{0}^{2} \sin αcos α}{g} = \frac{3}{4} \times \frac{\sqrt{20} \times \sqrt{20}}{10} = 1 .5 = 150 cm$