Question

A ball of mass$=100$gm is released from a height $h_1=2.5m$ from the ground level and then rebounds to a height $h_2=0.625m$ The time of contact of the ball and the ground is $\Delta t=0.01$ sec. The impulsive (impact) force offered by the ball on the ground is

• A. $105N$
• B. $1.50N$
• C. $2.08N$
• D. $208N$

Answer 1

The correct option is A $105N$

As we know that impulsive force is give by
F.t=m(V-U)
now we have to find the velocity before collision with ground
so $mg{h}_1$=$\frac{1}{2}m{v_1}^2$
$v_1^2$=$\sqrt{2g{h}_1}$
$v_1$=7m/sec
again the velocity of ball after collision with ground will be
$mg{h}_2$=$\frac{1}{2}m{v_2}^2$
$v_2^2$=$\sqrt{2g{h}_2}$
$v_2$=3.50m/sec
hence net velocity "V"=$v_1+v_2$=10.53m/sec
so from impulsive force equation we have to put all these value (U=0)
F=$\frac{m\left(V-U\right)}{t}$=$\frac{0.1\left(10.53\right)}{0.01}$=105.3 N