momentum and impulse - Two identical billiard balls strike a rigid wall with the same speed but at different angles, and get reflected without any change in speed, as shown in Fig. 5.6. What is (i) the direction of the force on the wall due to each ball? (ii) the ratio of the magnitudes of impulses imparted to the balls by the wall ?

Sandra (last edited 2 years ago) (Student, UG1 Grade)

(i) Direction of the force

Assume that the angle of incidence of ball 1 on the wall is = A.

Assume that the angle of incidence of the ball 2 is = B.

Let the speed of the balls just before they hit the wall equal v.

The incident and reflected velocity vectors are resolved along the standard to the wall and along the wall.

For the change in velocity of the ball 1 =v_f - v_i

= (v \cos A j + v \sin A i) - (v \sin A i - v cos A j)

= 2 v cos A j

Impulse = change in ball momentum 1 = 2 m v cos A j

So the force direction = momentum direction = natural to the wall, coming out of the wall.

Ratio of impulses = \frac{2 m v \cos A}{2 m v \cos B)}

= \frac{\cos A}{\cos B}

= \frac{\cos 30\degree}{\cos 30\degree}

= 1