at a ball game consider a baseball of mass m=0.15 that falls directly downward at a speed v=40m/s into the hands of a fan. what impulseft must be supplied to bring the ball to rest? if the ball is stopped in 0.3s, what is the average force of the ball on the catchers hand?
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To find the impulse required to bring the ball to rest, we can use the impulse-momentum principle, which states that the change in momentum of an object is equal to the impulse applied to it.
Impulse (J) is defined as the product of force (F) and time (Δt). Mathematically, it can be written as:
J = F * Δt
We can also calculate the change in momentum (Δp) of the ball using:
Δp = m * Δv
where m is the mass of the object and Δv is the change in velocity.
In this case, the mass of the baseball is given as m = 0.15 kg, and the initial velocity (v) is 40 m/s. Since the ball is brought to rest, the final velocity (v') is 0 m/s.
Therefore, the change in velocity Δv = v' - v = 0 - 40 = -40 m/s.
First, let's calculate the impulse (J):
J = F * Δt
Since we are given the stopping time Δt = 0.3 s, we can find the impulse:
J = F * 0.3
To find the average force (F) of the ball on the catcher's hand, we need to divide both sides of the equation by Δt:
F = J / Δt
Substituting the values:
F = (m * Δv) / Δt
F = (0.15 kg * -40 m/s) / 0.3 s
Now we can solve for F:
F = -20 N
The negative sign indicates that the force is in the opposite direction of the motion.
Therefore, to bring the ball to rest, an impulse of 6 N·s must be supplied, and the average force of the ball on the catcher's hand is -20 N.