Respuesta :
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,
[tex]Mv+mu=Mv_1+mv_2[/tex]
(900 x 47) + (200 x -30) = (900 x [tex]v_1[/tex]) + (200 x [tex]v_2[/tex])
36300 = (900 x [tex]v_1[/tex]) + (200 x [tex]v_2[/tex])
[tex]9v_1 + 2v_2 = 363[/tex] ..............(i)
[tex]9v_1 = 363 - 2v_2[/tex]
[tex]v_1=\frac{363 - 2v_2}{9}[/tex]
The mathematical expression for the conservation of kinetic energy is
[tex]\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2[/tex]
[tex]\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2[/tex] ................(ii)
[tex]$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$[/tex]
[tex]21681 = 9v_1^2+2v_2^2[/tex]
Substituting (i) in (ii)
[tex]21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2[/tex]
[tex](363-2v_2)^2+18v_2^2=195129[/tex]
[tex](363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0[/tex]
[tex]22v_2^2-145v_2-63360=0[/tex]
Solving the equation, we get
[tex]v_2=96 \ m/s, -30 \ m/s[/tex]
The negative velocity is neglected.
Therefore, substituting 96 m/s for [tex]v_2[/tex] in (i), we get
[tex]v_1=\frac{363-(2 \times 96)}{9}[/tex]
= 19
Thus, only impulse of importance is used to find final velocity.
