Question

A ball of mass 4 kg strikes a floor with a velocity of 10 m/s at an angle of 37° with the normal to the floor. The ball bounces off of the floor with the exact same speed along a path with the same angle. What is the impulse applied to the ball by the floor?

Answer 1

Given:

The mass of the ball is,

`m=4\text{ kg}`

The initial velocity of the ball is,

`v_i=10\text{ m/s}`

The angle with the floor is,

`\theta=37^(\circ)`

The change in momentum is the impulse by the floor, so the impulse is,

`\begin{gathered} P=m(v_f-v_i) \\ =4\lbrack10-(-10\rbrack \\ =80\text{ kg m/s} \end{gathered}`

The impulse is 80 kgm/s along with the final velocity.