Answer
(a) The racket is moving at a speed of 6.4 m/s after the collision.
(b) The average force exerted by the racket on the ball is 360 N
The racket's force is 610 times greater than the gravitational force on the ball.
Work Step by Step
(a) We first find the impulse $J$ that the racket exerted on the ball;
$p_0+J=p_f$
$J = p_f-p_0$
$J = m~(v_f-v_0)$
$J = (0.060~kg)[(40~m/s)-(-20~m/s)]$
$J = 3.6~N~s$
The ball also exerted the same magnitude of impulse on the racket in the opposite direction. Using this, we can find the speed of the racket after the collision:
$p_f = p_0+J$
$m~v_f = m~v_0 +J$
$v_f = \frac{m~v_0 +J}{m}$
$v_f = \frac{(1.0~kg)(10~m/s)-3.6~N~s}{1.0~kg}$
$v_f = 6.4~m/s$
The racket is moving at a speed of 6.4 m/s after the collision.
(b) We can use the impulse to find the average force exerted by the racket on the ball.
$F~t = J$
$F = \frac{J}{t}$
$F = \frac{3.6~N~s}{10\times 10^{-3}~s}$
$F = 360~N$
The average force exerted by the racket on the ball is 360 N.
We then find the gravitational force on the ball:
$F_g = m~g$
$F_g = (0.060~kg)(9.80~m/s^2)$
$F_g = 0.59~N$
We can find the ratio of the racket's force and the gravitational force on the ball.
$\frac{F}{F_g} = \frac{360~N}{0.59~N} = 610$
The racket's force is thus 610 times greater than the gravitational force on the ball.
