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.)

    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 =

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

         Δpₓ = m - 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)

          = 1.0686 N s

the total impulse is

          I = Iₓ i ^ + I_y j ^

          I = 1.06886  j^ N s

the independent variable, the one that is intentionally manipulated, is whether or not the car hits the barrier.

the dependent variable, the one that you measure the response in, is the distance that the washers travel.

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explanation: