When a 0.35 kg baseball is hit, its velocitychanges from 21 m/s to -29.5 m/s. What is the magnitude of the impulse de-livered by the bat to the ball?
Answer in units of N · s
32 N*s
It equals the momentum change, which is
50.5 x 0.35 N*s
You must be in Dr Jordans class as FAU as well
To find the magnitude of the impulse delivered by the bat to the ball, we first need to calculate the change in velocity.
The change in velocity (Δv) is given by the final velocity (vf) minus the initial velocity (vi). In this case, vf = -29.5 m/s and vi = 21 m/s.
Δv = vf - vi
Δv = -29.5 m/s - 21 m/s
Δv = -50.5 m/s
The impulse (J) is defined as the product of the force (F) applied to an object and the time (Δt) over which the force is applied.
J = F * Δt
Now, assuming the force applied by the bat to the ball is constant, we can rewrite the equation as J = F * Δt = m * Δv, where m is the mass of the baseball.
J = m * Δv
J = 0.35 kg * (-50.5 m/s)
J = -17.675 N · s
The magnitude of the impulse delivered by the bat to the ball is simply the absolute value of J, so:
|J| = |-17.675 N · s|
|J| = 17.675 N · s
Therefore, the magnitude of the impulse delivered by the bat to the ball is 17.675 N · s.