Question

A golf ball strikes a hard, smooth floor at an angle of 28.2 ° and, as the drawing shows, rebounds at the same angle. The mass of the ball is 0.0260 kg, and its speed is 42.8 m/s just before and after striking the floor. What is the magnitude of the impulse applied to the golf ball by the floor? (Hint: Note that only the vertical component of the ball's momentum changes during impact with the floor, and ignore the weight of the ball.)

Answer 1

Answer:

    I = 1.06886  N s

Explanation:

The expression for momentum is

          I = F t = Δp

therefore the momentum is a vector quantity, for which we define a reference system parallel to the floor

Let's find the components of the initial velocity

          sin 28.2 = v_y / v

          cos 28.2=  vₓ / v

          v_y = v sin 282

          vₓ = v cos 28.2

          v_y = 42.8 sin 28.2 = 20.225 m / s

          vₓ = 42.8 cos 28.2 = 37.72 m / s

since the ball is heading to the ground, the vertical velocity is negative and the horizontal velocity is positive, it can also be calculated by making

θ = -28.2

         v_y = -20.55 m / s

         v_x = 37.72 m / s

X axis

         Iₓ = Δpₓ = $p_{fx} - p_{ox}$

since the ball moves in the x-axis without changing the velocity, the change in moment must be zero

         Δpₓ = m $v_{fx}$ - m v₀ₓ = 0

          v_{fx} = v₀ₓ

therefore

          Iₓ = 0

Y axis  

        I_y = Δp_y = p_{fy} -p_{oy}

when the ball reaches the floor its vertical speed is downwards and when it leaves the floor its speed has the same modulus but the direction is upwards

         v_{fy} = - v_{oy}

         Δp_y = 2 m v_{oy}

         Δp_y = 2 0.0260 (20.55)

         $\Delta p_{y}$ = 1.0686 N s

the total impulse is

          I = Iₓ i ^ + I_y j ^

          I = 1.06886  j^ N s