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
mā = mass of clay ball traveling east = 28 g
vā = velocity of ball traveling east = 3.5 i m/s
mā = mass of clay ball traveling north Ā = 35 g
vā = 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
mā vā + mā vā = m v
(28) (3.5 i) + (35) (2.8 j ) = 63 v
98 i + 98 j Ā = 63 v
1.6 i + 1.6 j = v
Direction :
Īø = tanā»Ā¹(1.6/1.6)
Īø = 45 deg north of east
Speed is given as
|v| = sqrt((1.6)Ā² + (1.6)Ā²)
|v| = 2.3 m/s
