A 0.15kg baseball pitched at 30 m/s is hit on a horizontal line drive
straight back at the pitcher at 46.0 m/s. If the contact time between bat and ball is 5x10^-3 s, calculate the force (assumed to be constant) between the ball and bat. [Hint: impulse!]
I did FΔt = Δp = mΔv
for Δv I'm not sure if it is (30)-(46)=(-16) or (30)-(-46)=(76).
I think it is 76, so then I did:
(0.15)(76) = 11.4 kg *m/s
then 11.4/(5x10^-3) = 2280 N
a = (V-Vo)/t = (-46-30)/5*10^-3 =
-15.2*10^3 m/s^2.
F = m*a = 0.15 * (-15,200) = -2280 N.
To calculate the force between the ball and the bat, you correctly used the principle of impulse.
Impulse is defined as the change in momentum of an object and is given by the product of force and time: Δp = FΔt.
In this case, we can calculate impulse by subtracting the initial momentum from the final momentum.
Given that the mass of the baseball (m) is 0.15 kg, the initial velocity (vi) is 30 m/s, and the final velocity (vf) is -46 m/s (since it is hit back towards the pitcher), we can find the change in velocity (Δv) as follows:
Δv = vf - vi
= (-46) - 30
= -76 m/s
Note that the negative sign indicates the change in direction.
Now, using the impulse equation Δp = FΔt, we can substitute the values and calculate the force (F):
FΔt = Δp
F(5x10^-3 s) = (0.15 kg)(-76 m/s)
F = (0.15 kg)(-76 m/s) / (5x10^-3 s)
Evaluating this expression, we get:
F = -11.4 kg m/s / 5x10^-3 s
F = -2280 N
The negative sign indicates that the force is directed opposite to the motion of the ball.
Therefore, the force between the ball and the bat is 2280 N.