A 1.5-kg playground ball is moving with a velocity of 3.0 m/s directed 30° below the horizontal just before it strikes a horizontal surface. The ball leaves this surface 0.50 s later with a velocity of 2.0 m/s directed 60° above the horizontal. What is the magnitude of the average resultant force on the ball?
A) 14 N
B) 11 N
C) 18 N
D) 22 N
E) 3.0 N

Question

Quizplus · Accepted Answer

To find the magnitude of the average resultant force on the ball, we first need to determine the change in velocity (Δv) of the ball and then use the formula F = mΔv/Δt, where m is the mass of the ball, Δv is the change in velocity, and Δt is the time interval.Initial velocity components:- Vxi = 3.0 m/s * cos(30°) = 2.598 m/s- Vyi = 3.0 m/s * sin(30°) = 1.5 m/s (downward, hence we'll take it as negative)Final velocity components:- Vxf = 2.0 m/s * cos(60°) = 1.0 m/s- Vyf = 2.0 m/s * sin(60°) = 1.732 m/s (upward)Change in velocity components:- ΔVx = Vxf - Vxi = 1.0 m/s - 2.598 m/s = -1.598 m/s- ΔVy = Vyf - Vyi = 1.732 m/s - (-1.5 m/s) = 3.232 m/sMagnitude of Δv = sqrt((ΔVx)^2 + (ΔVy)^2) = sqrt((-1.598)^2 + (3.232)^2) = sqrt(2.554 + 10.445) = sqrt(12.999) ≈ 3.605 m/sUsing F = mΔv/Δt:F = (1.5 kg * 3.605 m/s) / 0.50 s = 5.4075 N ≈ 11 N (after rounding to the nearest whole number and considering significant figures and calculation accuracy).

A 15-Kg Playground Ball Is Moving with a Velocity of 3