Answer:
The maximum velocity is 1.58 m/s.
Explanation:
A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.
Spring constant, K = 100 N/m
mass, m = 0.1 kg
Amplitude, A = 5 cm = 0.05 m
Let the angular frequency is w.
The maximum velocity is
Incomplete question as the mass of baseball is missing.I have assume 0.2kg mass of baseball.So complete question is:
A baseball has mass 0.2 kg.If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
Answer:
ΔP=20 kg.m/s
Explanation:
Given data
Mass m=0.2 kg
Initial speed Vi=-44.5m/s
Final speed Vf=55.5 m/s
Required
Change in momentum ΔP
Solution
First we take the batted balls velocity as the final velocity and its direction is the positive direction and we take the pitched balls velocity as the initial velocity and so its direction will be negative direction.So we have:
Now we need to find the initial momentum
So
Substitute the given values
Now for final momentum
So the change in momentum is given as:
ΔP=P₂-P₁
ΔP=20 kg.m/s