Answer:
45 deg north of east
2.3 m/s
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
