A 0.166 kg baseball, thrown with a speed of 38.6 m/s, is hit straight back at the pitcher with a speed of 47.9 m/s.
(a) What is the magnitude of the impulse delivered by the bat to the baseball?
answer: 14.4 kg*m/s
(b) Find the magnitude of the average force exerted by the bat on the ball if the two are in contact for 2.15 10-3 s. (I just need help figuring out part b)
(b) f = m a = .166 kg * [(47.9 m/s + 38.6 m/s) / 2.15E-3 s]
To find the magnitude of the average force exerted by the bat on the ball, we can use the impulse-momentum theorem, which states that the impulse experienced by an object is equal to its change in momentum.
In equation form, the impulse (J) is given by the equation:
J = Δp = m * Δv
Where:
J = impulse
Δp = change in momentum
m = mass of the object
Δv = change in velocity
For part (a), we already have the initial and final velocities of the baseball. The change in velocity (Δv) can be calculated by subtracting the initial velocity from the final velocity:
Δv = vf - vi
Substituting the given values:
Δv = 47.9 m/s - (-38.6 m/s) = 86.5 m/s
Now, using the equation J = m * Δv, we can calculate the impulse:
J = (0.166 kg) * (86.5 m/s) = 14.4 kg*m/s
Now, moving on to part (b), we need to find the average force exerted by the bat on the ball during the time they are in contact. The relationship between impulse and average force is given by the equation:
J = F * Δt
Where:
F = average force
Δt = time interval
Rearranging the equation, we get:
F = J / Δt
Substituting the known values:
F = 14.4 kg*m/s / (2.15 * 10^-3 s) ≈ 6702 N
Therefore, the magnitude of the average force exerted by the bat on the ball is approximately 6702 N.