Respuesta :
Answer:
[tex]v_r=13.68\ m.s^{-1}[/tex] is the final velocity of the racket.
[tex]F=18.86\ N[/tex]
Explanation:
Given:
- mass of the racket, [tex]m_r=1000\ g[/tex]
- mass of ball, [tex]m_b=60\ g[/tex]
- initial speed of racket, [tex]u_r=15\ m.s^{-1}[/tex]
- initial speed of ball, [tex]u_b=18\ m.s^{-1}[/tex]
- final speed of ball, [tex]v_b=40\ m.s^{-1}[/tex]
- time of contact of racket with the ball, [tex]t=0.07\ s[/tex]
By the law of conservation of momentum:
[tex]m_r.u_r+m_b.u_b=m_r.v_r+m_b.v_b[/tex]
where: [tex]v_r=[/tex] final velocity of the racket
[tex]1000\times 15+60\times 18=1000\times v_r+60\times 40[/tex]
[tex]v_r=13.68\ m.s^{-1}[/tex] is the final velocity of the racket.
By the Newton's second law of motion:
[tex]F=\frac{dp}{dt}[/tex] ............................(1)
where:
dp = change in momentum
dt = change in time
Change in momentum of ball:
[tex]\Delta p_b=m_b.v_b-m_b.u_b[/tex]
[tex]\Delta p_b=60\times 10^{-3}\times (40-18)[/tex]
[tex]\Delta p_b=1.32\ kg.m.s^{-1}[/tex]
Now, using eq.(1):
[tex]F=\frac{1.32}{0.07}[/tex]
[tex]F=18.86\ N[/tex]
