Respuesta :
The impulse exerted over an object is equal to the change in its linear momentum:
[tex]I=\Delta p[/tex]On the other hand, the impulse is equal to the force exerted over the object multiplied by the time during which the force was exerted:
[tex]I=F\cdot\Delta t[/tex]First, find the impulse by finding the change in linear momentum of the baseball. The linear momentum is given by the product of the speed of the baseball times its mass:
[tex]p=mv[/tex]Assume that the negative direction is towards the batter and the positive direction is towards the pitcher. Then, the initial velocity of the ball is -42.3 m/s and the final velocity of the ball is 31.0 m/s. Then:
[tex]\begin{gathered} p_i=mv_i=(0.140\operatorname{kg})(-42.3\frac{m}{s})=-5.922\operatorname{kg}\cdot\frac{m}{s} \\ \\ p_f=mv_f=(0.140\operatorname{kg})(31.0\cdot\frac{m}{s})=4.34\operatorname{kg}\cdot\frac{m}{s} \end{gathered}[/tex]Use the initial and final linear momentum to find the change in linar momentum, which is equal to the impulse:
[tex]\begin{gathered} I=\Delta p \\ =p_f-p_i \\ =(4.34\operatorname{kg}\cdot\frac{m}{s})-(-5.922\operatorname{kg}\cdot\frac{m}{s}) \\ =10.262\operatorname{kg}\cdot\frac{m}{s} \end{gathered}[/tex]Isolate Δt from the equation that relates force and impulse and substitute the corresponding values for I and F to find the time during which the bat and the ball were in contact:
[tex]\begin{gathered} \Delta t=\frac{I}{F} \\ =\frac{10.262\operatorname{kg}\cdot\frac{m}{s}}{5120N} \\ =0.00200429687\ldots s \\ \approx0.00200s=2.00ms \end{gathered}[/tex]Therefore, the bat and the ball were in contact during a time interval of 2 miliseconds.
