Question

A 28 g ball of clay traveling east at 3.5 m/s collides with a 35 g ball of clay traveling north at 2.8 m/s.

What is the movement direction of the resulting 63 g blob of clay?

What is the speed of the resulting 63 g blob of clay?

Answer 1

45 deg north of east

2.3 m/s

Step-by-step explanation:

Assuming east as directed along positive x-axis and north as directed along positive y-axis

m₁ = mass of clay ball traveling east = 28 g

v₁ = velocity of ball traveling east = 3.5 i m/s

m₂ = mass of clay ball traveling north = 35 g

v₂ = velocity of ball traveling north = 2.8 j m/s

m = mass of the combination of two balls after collision = 63 g

v = velocity of the combination after collision

Using conservation of momentum

m₁ v₁ + m₂ v₂ = m v

(28) (3.5 i) + (35) (2.8 j ) = 63 v

98 i + 98 j = 63 v

1.6 i + 1.6 j = v

Direction :

θ = tan⁻¹(1.6/1.6)

θ = 45 deg north of east

Speed is given as

|v| = sqrt((1.6)² + (1.6)²)

|v| = 2.3 m/s