A softball (350g) is thrown with an initial speed of 27m/s. The moment just before the softball hits the ground, VelocityFx = +27m/s and VelocityFy = -36m/s. Determine the final momentum of the softball.
Momentum is a vector. Add Mx and My as vectors
finalmomentum=sqrt(Mx^2+My^2)
To determine the final momentum of the softball, we first need to find the final velocity.
Given:
Initial velocity in the x-direction (Vix) = 27 m/s
Final velocity in the x-direction (Vfx) = 27 m/s
Final velocity in the y-direction (Vfy) = -36 m/s
Since the final velocity in the x-direction remains the same as the initial velocity, we can conclude that the x-component of the velocity does not change during the motion. Therefore, the x-component of the final velocity (Vfx) is 27 m/s.
To find the magnitude of the final velocity (Vf), we can use the Pythagorean theorem:
Vf = √(Vfx² + Vfy²)
= √(27² + (-36)²)
= √(729 + 1296)
= √2025
= 45 m/s
Now that we know the magnitude of the final velocity, we can calculate the final momentum (Pf) using the equation:
Pf = mass × velocity
= 0.350 kg × 45 m/s
= 15.75 kg·m/s
Therefore, the final momentum of the softball is 15.75 kg·m/s.