Respuesta :
Answer:
Part(a): The expression for the velocity of the billiard ball is [tex]\bf{v_{2f} = \dfrac{2m_{1}v}{m_{1}+m_{2}}}[/tex]
Part(b): The value of the velocity of the billiard ball is [tex]\bf{5.57~m/s}[/tex].
Part(c): The expression for the velocity of the cue ball is [tex]\bf{v_{1f} = \dfrac{(m_{1} - m_{2})v}{m_{1}+m_{2}}}[/tex]
Part(d): The value of the velocity of the cue ball is [tex]\bf{1.93~m/s}[/tex].
Explanation:
Given:
The mass of the cue ball, [tex]m_{1} = 0.34~kg[/tex].
The mass of the billiard ball, [tex]m_{2} = 0.575~kg[/tex].
The initial velocity of the cue ball, [tex]u_{1} = 7.5~m/s[/tex]
The initial velocity of the billiard ball, [tex]u_{2} = 0[/tex]
(a)
Consider the final velocity of the cue ball be [tex]v_{1}[/tex] and the final velocity of the billiard ball be [tex]v_{2}[/tex].
From the conservation of linear momentum , we can write
[tex]~~~~&& m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}\\&or,& m_{1}(u_{1} - v_{1}) = m_{2}(v_{2} - u_{2})~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]
From the conservation of energy, we can write
[tex]~~~&& \dfrac{1}{2}m_{1}u_{1}^{2} + \dfrac{1}{2}m_{2}u_{2}^{2} = \dfrac{1}{2}m_{1}v_{1}^{2} + \dfrac{1}{2}m_{2}v_{2}^{2}\\&or,& \dfrac{1}{2}m_{1}(u_{1}^{2} - v_{1}^{2}) = \dfrac{1}{2}m_{2}(v_{2}^{2} - u_{2}^{2})~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]
Dividing equation (1) by equation (2), we have
[tex]~~~&& \dfrac{m_{1}(u_{1} - v_{1}) }{m_{1}(u_{1}^{2} - v_{1}^{2})} = \dfrac{m_{2}(v_{2} - u_{2})}{m_{2}(v_{2}^{2} - u_{2}^{2})}\\&or,& u_{1} + v_{1} = u_{2} + v_{2}~~~~~~~~~~~~~~~~~~~~~~~~(3)[/tex]
Rearranging equation (3) for [tex]v_{1}[/tex], we have
[tex]v_{1} = u_{2} + v_{2} - u_{1}~~~~~~~~~~~~~~~~~~~~(4)[/tex]
Substitute equation (4) in equation (1), we can write
[tex]v_{2} = \dfrac{2m_{1}u_{1}}{m_{1}+m_{2}} + \dfrac{m_{2} - m_{1}}{m_{1} + m_{2}}u_{2}~~~~~~~~~~~~~~(5)[/tex]
Substituting [tex]0[/tex] for [tex]u_{2}[/tex], [tex]v[/tex] for [tex]u_{1}[/tex] and [tex]v_{2f}[/tex] for [tex]v_{2}[/tex] in equation (5), we have
[tex]v_{2f} = \dfrac{2m_{1}v}{m_{1}+m_{2}}~~~~~~~~~~~~~~~~~~~~~~~~(6)[/tex]
(b)
Substituting [tex]0.34~kg[/tex] for [tex]m_{1}[/tex], [tex]7.5~m/s[/tex] for [tex]v[/tex] and [tex]0.575~kg[/tex] for [tex]m_{2}[/tex] in equation (6), we have
[tex]v_{2f} &=& \dfrac{2(0.34~kg)(7.5~m/s)}{(0.34 + 0.575)~kg}\\~~~~~&=& 5.57~m/s[/tex]
(c)
Rearranging equation(3) for [tex]v_{2}[/tex], we have
[tex]v_{2} = u_{1} + v_{1} – u_{2}~~~~~~~~~~~~~~~~~~~~(7)[/tex]
Substitute equation (7) in equation (1), we can write
[tex]v_{1} = \dfrac{2m_{2}u_{2}}{m_{1}+m_{2}} + \dfrac{m_{1} - m_{2}}{m_{1} + m_{2}}u_{2}~~~~~~~~~~~~~~(8)[/tex]
Substituting [tex]0[/tex] for [tex]u_{2}[/tex], [tex]v[/tex] for [tex]u_{1}[/tex] and [tex]v_{1f}[/tex] for [tex]v_{1}[/tex] in equation (8), we have
[tex]v_{1f} = \dfrac{(m_{1} - m_{2})v}{m_{1}+m_{2}}~~~~~~~~~~~~~~~~~~~~~~~~(9)[/tex]
(d)
Substituting [tex]0.34~kg[/tex] for [tex]m_{1}[/tex], [tex]7.5~m/s[/tex] for [tex]v[/tex] and [tex]0.575~kg[/tex] for [tex]m_{2}[/tex] in equation (9), we have
[tex]v_{1f} &=& \dfrac{(0.34 - 0.575)~kg(7.5~m/s)}{(0.34 + 0.575)~kg}\\~~~~~&=& -1.93~m/s[/tex]
Negative sign indicates that the cue ball will bounce back.
