Respuesta :
Answer:
a) [tex]v=8.564\ m.s^{-1}[/tex]
b) [tex]\Delta KE=45.76\ J[/tex]
c) [tex]v=8.358\ m.s^{-1}[/tex]
d) [tex]\Delta KE=225.24\ J[/tex]
Explanation:
Given:
mass of the player, [tex]m_p=102.5\ kg[/tex]
mass of the ball, [tex]m_b=0.47\ kg[/tex]
initial velocity of the player, [tex]v_p=8.5\ m.s^{-1}[/tex]
initial velocity of the ball, [tex]v_b=22.5\ m.s^{-1}[/tex]
a)
Case: When the player and the ball are moving in the same direction.
[tex]m_t.v=m_p.v_p+m_b.v_b[/tex]
where:
[tex]m_t=[/tex]total mass after the player catches the ball
v = final velocity of the system
[tex]v=\frac{102.5\times 8.5+0.47\times 22.5}{(102.5+0.47)}[/tex]
[tex]v=8.564\ m.s^{-1}[/tex]
b)
Initial kinetic energy of the system:
[tex]KE_i=\frac{1}{2} [m_p.v_p^2+m_b.v_b^2][/tex]
[tex]KE_i=\frac{1}{2} [102.5\times 8.5^2+0.47\times 22.5^2][/tex]
[tex]KE_i=3821.78\ J[/tex]
Final kinetic energy of the system:
[tex]KE_f=\frac{1}{2} m_t.v^2[/tex]
[tex]KE_f=\frac{1}{2}\times 102.97\times 8.564^2[/tex]
[tex]KE_f=3776.02\ J[/tex]
∴Change in kinetic energy
[tex]\Delta KE=KE_i-KE_f[/tex]
[tex]\Delta KE=3821.78-3776.02 [/tex]
[tex]\Delta KE=45.76\ J[/tex]
c)
Case: When the player and the ball are moving in the opposite direction.
[tex]m_t.v=m_p.v_p-m_b.v_b[/tex]
[tex]v=\frac{102.5\times 8.5-0.47\times 22.5}{(102.5+0.47)}[/tex]
[tex]v=8.358\ m.s^{-1}[/tex]
d)
Final kinetic energy in this case:
[tex]KE_f=\frac{1}{2} m_t.v^2[/tex]
[tex]KE_f=0.5\times 102.97\times 8.358^2[/tex]
[tex]KE_f=3596.54\ J[/tex]
∴Change in kinetic energy:
[tex]\Delta KE=KE_i-KE_f[/tex]
[tex]\Delta KE=3821.78-3596.54[/tex]
[tex]\Delta KE=225.24\ J[/tex]
