Question

A 3.02 kg object initially moving in the positive x direction with a velocity of 4.63 m s collides with and sticks to a 2.14 kg object initially moving in the negative y direction with a velocity of 2.99 m s Find the final components of velocity of the composite object Indicate the direction with the sign of your answer

Answer 1

Answer:

v = (2.71î + 1.24j) m/s

Magnitude of v = 2.98 m/s

Direction of v = 24.6°

Explanation:

Let the final velocity of the composite of both objects after collision be v.

According to Newton's second law of motion, in collisions, momentum is normally conserved

Momentum before collision = Momentum after collision

Momentum of object 1 before collision = (3.02)(4.63î) = (13.9826 î) kgm/s

Momentum of object 2 before collision = (2.14)(2.99j) = (6.3986 j) kgm/s

Momentum of both objects after collision = (3.02 + 2.14)(v) = 5.16v

Momentum balance

(13.9826î + 6.3986j) = 5.16v

v = (2.71î + 1.24j) m/s

Magnitide = √(2.71² + 1.24²)

Magnitude of v = 2.98 m/s

Direction of v = tan⁻¹ (1.24/2.71)

Direction of v = 24.6°

Hope this Helps!!!