Answer:
[tex]\mathbf{^\to Q =0.84375 \ (- \hat z) \ kgm^2s^{-1}}[/tex]
Explanation:
mass of the ball = 0.25 kg
its speed v = 4.5 m/s
r = 0.75 m
let the speed be in the horizontal direction and the distance r be in the vertical direction we have :
[tex]^ {\to}v = 4.5 \ x \ \ m/s[/tex]
[tex]^ {\to}r = 0.75 \ y \ \ m/s[/tex]
Let the momentum about the center of intersection be P;
SO;
[tex]^ \to} P = ^{\to} mv[/tex]
[tex]^ \to} P = 0.25* 4.5 \ x \ \ m/s[/tex]
[tex]^ \to} P = 1.125 \ x \ \ m/s[/tex]
Let the angular momentum be Q;
[tex]^\to Q = ^ \to r* ^ \to P[/tex]
[tex]^\to Q = (0.75*1.125) kgm^2s^{-1} *(\hat y * \hat x)[/tex]
[tex]\mathbf{^\to Q =0.84375 \ (- \hat z) \ kgm^2s^{-1}}[/tex]
