A(n) 0.42 kg softball is pitched at a speed of 12 m/s. The batter hits it back directly at the pitcher at a speed of 27 m/s. The bat acts on the ball for 0.016 s. What is the magnitude of the impulse imparted by the bat to the ball? Answer in units of N · s.
What is the magnitude of the average force exerted by the bat on the ball? Answer in units of N.
impulse= changemomentum=39*.42 N s
force= impulse/time
To find the magnitude of the impulse imparted by the bat to the ball, we can use the equation:
Impulse = Change in momentum
The momentum of an object is given by the equation:
Momentum = mass * velocity
Given:
Mass of the ball (m) = 0.42 kg
Initial velocity of the ball (v1) = 12 m/s
Final velocity of the ball (v2) = -27 m/s (negative because it is in the opposite direction)
Since we know the initial and final velocities, we can determine the change in velocity:
Change in velocity (Δv) = v2 - v1 = -27 m/s - 12 m/s = -39 m/s
Now, we can calculate the impulse:
Impulse = mass * Δv = 0.42 kg * (-39 m/s) = -16.38 N · s
However, impulse is a vector quantity, so we need to consider its magnitude. Therefore, we take the absolute value:
Magnitude of Impulse = |-16.38 N · s| = 16.38 N · s
For the magnitude of the average force exerted by the bat on the ball, we can use the equation:
Average Force = Impulse / Time
Given:
Time (Δt) = 0.016 s
Average Force = 16.38 N · s / 0.016 s = 1023.75 N
So, the magnitude of the average force exerted by the bat on the ball is 1023.75 N.