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

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? 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?
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?

Explanation / Answer

An inelastic collsion:

m1v1 + m2v2 = V(m1 + m2) [Conservation of momentum]

East-West component:
(28 g)(3.5 m/s) + (35 g)(0 m/s) = V1(28 g + 35 g)
V1 = 1.556 m/s

North-south component:
(28 g)(0 m/s) + (35 g)(2.8 m/s) = V2(28 g + 35 g)
V2 = 1.556 m/s

By Pythagorean's Theorem:
V = sqrt(V1^2 + V2^2)
V = sqrt[(1.556 m/s)^2 + (1.556 m/s)^2]
V = 2.20 m/s

tan x = (1.556 m/s) / (1.556 m/s)
x = 45 degrees

Therefore, the 63 g clay ball is traveling at 2.20 m/s 45 degrees north of east.

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