Question

A 0.5 kg ball free fall from a height of $h_1$ = 7.2 m and reflected a height of $h_2$ = 3.2 m. Acceleration due to gravity = 10 $m/s^2$. Determine impulse.

Answer 1

Known :

Mass of ball (m) = 0.5 kg
First height (h1) = 7.2 m
Second height (h2) = 3.2 m
Acceleration due to gravity (g) = 10 $m/s^2$

Calculation for velocity of ball just before collision($v_0$):
$v^2$ = 2 g h
$v_o^2$ = 2(10)(7.2) = 144
$v_o$ = -12 m/s

Velocity of ball before collision (vo) = -12 m/s. (Negative sign indicates the downward direction of ball)

Calculation for velocity of ball just after collision($v_f$):
$v_t^2$ = $v_f^2$ + 2 g h
0 = $v_f^2$ + 2 (-10)(3.2)
$v_f^2$ = 64
$v_f$ = +8 m/s
Velocity of ball after collision ($v_f$) is 8 m/s

Impulse (I) = the change in momentum (Δp)

I = m ($v_f–v_0$) = (0.5)(8-(-12)) = (0.5)(8 + 12) = (0.5)(20) = 10 Ns