Solution :
Given :
Mass of the baseball, m = 200 g
Velocity of the baseball, u = -30 m/s
Mass of the baseball after struck by the bat, M = 900 g
Velocity of the baseball after struck by the bat, v = 47 m/s
According to the conservation of momentum,
![Mv+mu=Mv_1+mv_2](https://tex.z-dn.net/?f=Mv%2Bmu%3DMv_1%2Bmv_2)
(900 x 47) + (200 x -30) = (900 x
) + (200 x
)
36300 = (900 x
) + (200 x
)
..............(i)
![9v_1 = 363 - 2v_2](https://tex.z-dn.net/?f=9v_1%20%3D%20363%20-%202v_2)
![v_1=\frac{363 - 2v_2}{9}](https://tex.z-dn.net/?f=v_1%3D%5Cfrac%7B363%20-%202v_2%7D%7B9%7D)
The mathematical expression for the conservation of kinetic energy is
![\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7DMv%5E2%2B%5Cfrac%7B1%7D%7B2%7Dmu%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7DMv_1%5E2%2B%5Cfrac%7B1%7D%7B2%7Dmv_2%5E2)
................(ii)
![21681 = 9v_1^2+2v_2^2](https://tex.z-dn.net/?f=21681%20%3D%209v_1%5E2%2B2v_2%5E2)
Substituting (i) in (ii)
![21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2](https://tex.z-dn.net/?f=21681%3D%209%5Cleft%28%20%5Cfrac%7B363-2v_2%7D%7B9%7D%5Cright%29%5E2%2B2v_2%5E2)
![(363-2v_2)^2+18v_2^2=195129](https://tex.z-dn.net/?f=%28363-2v_2%29%5E2%2B18v_2%5E2%3D195129)
![(363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0](https://tex.z-dn.net/?f=%28363%29%5E2%2B18v_2%5E2-2%28363%29%282v_2%29%2B%28363%29%5E2-195129%3D0)
![22v_2^2-145v_2-63360=0](https://tex.z-dn.net/?f=22v_2%5E2-145v_2-63360%3D0)
Solving the equation, we get
![v_2=96 \ m/s, -30 \ m/s](https://tex.z-dn.net/?f=v_2%3D96%20%5C%20m%2Fs%2C%20-30%20%5C%20m%2Fs)
The negative velocity is neglected.
Therefore, substituting 96 m/s for
in (i), we get
![v_1=\frac{363-(2 \times 96)}{9}](https://tex.z-dn.net/?f=v_1%3D%5Cfrac%7B363-%282%20%5Ctimes%2096%29%7D%7B9%7D)
= 19
Thus, only impulse of importance is used to find final velocity.