Question

Suppose that a 102.5 kg football player running at 8.5 m/s catches a 0.47 kg ball moving at a speed of 22.5 m/s with his feet off the ground, while both of them are moving horizontally.a. calculate the final speed of the player in m/s, if the ball and the player are initiallly moving in the same direction
b.calculate the change in kinetic energy of the system, in joules, after the player catches the ball
c. calculate the final speed of the player in m/s, if the ball and the player are initially moving in opposite directions
d. calculate the change in kinetic energy of the system, in joules, in this case (opposite direction)

Answer 1

Answer:

a) $v=8.564\ m.s^{-1}$

b) $\Delta KE=45.76\ J$

c) $v=8.358\ m.s^{-1}$

d) $\Delta KE=225.24\ J$

Explanation:

Given:

mass of the player, $m_p=102.5\ kg$

mass of the ball, $m_b=0.47\ kg$

initial velocity of the player, $v_p=8.5\ m.s^{-1}$

initial velocity of the ball, $v_b=22.5\ m.s^{-1}$

a)

Case: When the player and the ball are moving in the same direction.

$m_t.v=m_p.v_p+m_b.v_b$

where:

$m_t=$total mass after the player catches the ball

v = final velocity of the system

$v=\frac{102.5\times 8.5+0.47\times 22.5}{(102.5+0.47)}$

$v=8.564\ m.s^{-1}$

b)

Initial kinetic energy of the system:

$KE_i=\frac{1}{2} [m_p.v_p^2+m_b.v_b^2]$

$KE_i=\frac{1}{2} [102.5\times 8.5^2+0.47\times 22.5^2]$

$KE_i=3821.78\ J$

Final kinetic energy of the system:

$KE_f=\frac{1}{2} m_t.v^2$

$KE_f=\frac{1}{2}\times 102.97\times 8.564^2$

$KE_f=3776.02\ J$

∴Change in kinetic energy

$\Delta KE=KE_i-KE_f$

$\Delta KE=3821.78-3776.02$

$\Delta KE=45.76\ J$

c)

Case: When the player and the ball are moving in the opposite direction.

$m_t.v=m_p.v_p-m_b.v_b$

$v=\frac{102.5\times 8.5-0.47\times 22.5}{(102.5+0.47)}$

$v=8.358\ m.s^{-1}$

d)

Final kinetic energy in this case:

$KE_f=\frac{1}{2} m_t.v^2$

$KE_f=0.5\times 102.97\times 8.358^2$

$KE_f=3596.54\ J$

∴Change in kinetic energy:

$\Delta KE=KE_i-KE_f$

$\Delta KE=3821.78-3596.54$

$\Delta KE=225.24\ J$