Question

A 0.25 ball is thrown with the speed of v=4.5m/s at a resting stick that can rotate around one end at the axis O. The ball hits the stick at r=0.75 m from O. What is the angular momentum of the ball about the axis O before hitting the stick ?

Answer 1

-0.84

Step-by-step explanation:

If you multiply 0.25 by 4 you get 1.

it's negative because of the mass so -1.

- 1 + 0.16 = -0.84

Answer 2

`\mathbf{^\to Q =0.84375 \ (- \hat z) \ kgm^2s^(-1)}`

Step-by-step 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 :

`^ {\to}v = 4.5 \ x \ \ m/s`

`^ {\to}r = 0.75 \ y \ \ m/s`

Let the momentum about the center of intersection be P;

SO;

`^ \to} P = ^(\to) mv`

`^ \to} P = 0.25* 4.5 \ x \ \ m/s`

`^ \to} P = 1.125 \ x \ \ m/s`

Let the angular momentum be Q;

`^\to Q = ^ \to r* ^ \to P`

`^\to Q = (0.75*1.125) kgm^2s^(-1) *(\hat y * \hat x)`

`\mathbf{^\to Q =0.84375 \ (- \hat z) \ kgm^2s^(-1)}`