42.) A 0.050 kg whildle ball is thrown hori zantally batter of tow ord The boter 12.3 mls a a the bat and hits the ballo suings an ave cage force f 15.714 N to the ball applying the ball to head back in the fausing diredion Cauou from the beather) at a.70 mis, Lossumpg e opposte away Rrom botter is the posstive determine the time that in Centact, Report that the bot and balll uere i hendretns of o ansaer your second ente humerical volve

In this case, an impulse is acted on the ball, which is the case of the change of momentum of the ball in a short time interval due to a large force acting on it.

The impulse on the ball can be found by the change in the momentum of the ball, as

$$\begin{gathered} I=p_2-p_1 \\ {p}_2 is the final and p_{1 }is the initial momentum of the ball \\ As the ball travels in the opposite direction after being hit, so the momenta of the ball are oppositely directed \\ I=p_2-(-p_1) \\ I=p_2+p_1 \\ I=m{v}_2+m{v}_1 \\ I=m(v_2+v_1) \\ I=0.05(9.7+12.3) \\ I=1.1 Ns \end{gathered}$$

This is the impulse acting on the ball

The impulse is also defined as the amount of force acting on the ball in a short time interval t.

So, impulse can also be written as

$$\begin{gathered} I=F∆t \\ ∆t is the time interval for which the force has been acted upon. \\ So, ∆t=\frac{I}{F} \\ ∆t=\frac{1.1}{15.714} \\ ∆t=0.07 s \\ ∆t=7\times {10}^{-2} s \end{gathered}$$

This is the time for which the bat and ball were in contact.