Question

A 20 g ball of clay traveling east at 3.5 m/s collides with a 35 g ball of clay traveling north at 2.0 m/s .

Part A What is the movement direction of the resulting 55 g blob of clay?

Express your answer in degrees north of east. θ θ =

Part B What is the speed of the resulting 55 g blob of clay?

Answer 1

Let us consider the east direction as the positive X-direction and the north direction as the positive Y-direction.

Mass of the first ball of clay = m₁ = 20 g = 0.02 kg

Mass of the second ball of clay = m₂ = 35 g = 0.035 kg

Initial velocity of the first ball of clay = V₁ = 3.5 m/s to the east

Initial velocity of the second ball of clay = V₂ = 2 m/s to the north

Final velocity of the blob of clay = V₃

X-component of the final velocity of the blob of clay = V_3x

Y-component of the final velocity of the blob of clay = V_3y

By conservation of linear momentum in the X-direction,

m₁V₁ = (m₁ + m₂)V_3x

(0.02)(3.5) = (0.02 + 0.035)V_3x

V_3x = 1.273 m/s

By conservation of linear momentum in the Y-direction,

m₂V₂ = (m₁ + m₂)V_3y

(0.035)(2) = (0.02 + 0.035)V_3y

V_3y = 1.273 m/s

V₃ = 1.8 m/s

Direction of the final velocity of the blob of clay =

= 45^o north of east

A) Movement direction of the resulting 55 g blob of clay = 45^o north of east

B) Speed of the resulting 55 g blob of clay = 1.8 m/s