Answer:
Velocity is 3.11 m/s at an angle of -56° with respect to the original line of motion.
Explanation:
Let line of action be horizontal axis , mass of ball be m and unknown velocity be v.
Here momentum is conserved.
Initial momentum =Final momentum
Initial momentum = m x 6i + m x 0i = 6m i
Final momentum = m x (5.21cos 29.7 i + 5.21sin 29.7 j) + m x v = 4.26 m i + 2.58 m j + m v
4.26 m i + 2.58 m j + m v = 6m i
v = 1.74 i - 2.58 j
Magnitude of velocity [tex]=\sqrt{1.74^2+(-2.58)^2}=3.11m/s[/tex]
Direction,
[tex]\theta =tan^{-1}\left ( \frac{-2.58}{1.74}\right )=--56^0[/tex]
Velocity is 3.11 m/s at an angle of -56° with respect to the original line of motion.
