Question

A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). The ball is initially traveling at speed V0 perpendicular to the stick. The ball strikes the stick at a distance d from its center. The collision is elastic.

a). Find the resulting translational and rotational speeds of the stick.
b). Find the resulting speed of the ball.

Answer 1

Part a)

`v_2 = ((2\beta mL^2v_o)/(d))/((md + (\beta mL^2)/(d)(1 + (m)/(M)))`

Part b)

`v_1 = v_0 - (m)/(M)(((2\beta mL^2v_o)/(d))/((md + (\beta mL^2)/(d)(1 + (m)/(M))))`

Step-by-step explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

`Mv_o = M v_1 + m v_2`

here we also use angular momentum conservation

so we have

`M v_o d = M v_1 d + \beta mL^2 \omega`

also we know that the collision is elastic collision so we have

`v_o = (v_2 + d\omega) - v_1`

so we have

`\omega = (v_o + v_1 - v_2)/(d)`

also we know

`M v_o d - M v_1 d = \beta mL^2((v_o + v_1 - v_2)/(d))`

also we know

`v_1 = v_o - (m)/(M)v_2`

so we have

`M v_o d - M(v_o - (m)/(M)v_2)d = \beta mL^2((v_o + v_o - (m)/(M)v_2 - v_2)/(d))`

`mv_2 d = \beta mL^2(2v_o)/(d) - \beta mL^2(1 + (m)/(M))(v_2)/(d)`

now we have

`(md + (\beta mL^2)/(d)(1 + (m)/(M))v_2 = (2\beta mL^2v_o)/(d)`

`v_2 = ((2\beta mL^2v_o)/(d))/((md + (\beta mL^2)/(d)(1 + (m)/(M)))`

Part b)

Now we know that speed of the ball after collision is given as

`v_1 = v_o - (m)/(M)v_2`

so it is given as

`v_1 = v_0 - (m)/(M)(((2\beta mL^2v_o)/(d))/((md + (\beta mL^2)/(d)(1 + (m)/(M))))`