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?
Impulse = change of momentum
= m(v0-v1)
Average force = change of momentum / time
= m(v0-v1)/0.3 N (assuming m=0.15 kg)
To find the impulse, we can use the equation:
Impulse (J) = change in momentum (Δp)
The momentum of an object is given by:
Momentum (p) = mass (m) * velocity (v)
In this case, the mass of the baseball (m) is given as 0.15 kg, and the velocity (v) is given as 40 m/s. Thus, the initial momentum (p_initial) of the baseball can be calculated as:
p_initial = m * v
To bring the ball to rest, the final momentum (p_final) of the ball is 0 kg m/s, since it comes to a stop. Therefore, the change in momentum (Δp) is:
Δp = p_final - p_initial
Substituting the values, we get:
Δp = 0 - (m * v)
Now, to find the impulse (J), we multiply the change in momentum (Δp) by the time (t) during which the change occurs, given by:
J = Δp * t
In this case, the time (t) is given as 0.3 seconds. Substituting the values, we have:
J = (0 - (m * v)) * t
Now we can calculate the impulse (J) using the given values of mass (m) and velocity (v), as well as the time (t).
Next, to find the average force (F_avg) exerted by the ball on the catcher's hand, we can use the equation:
F_avg = J / t
Again, substituting the values, we have:
F_avg = J / t
By plugging the previously calculated impulse (J) and the given time (t) into this equation, we can calculate the average force (F_avg) exerted by the ball on the catcher's hand.