Question

A ball of mass $200$g falls of the ground from a height of $2.5$m and rises to a height of $0.4$m on hitting the ground. If the time of contact of the hall with the ground is $0.02$s, what is the force exerted by the ground on the ball? ($g=9.8m/s^2$)

• A. $111.9N$
• B. $98.9N$
• C. $125.6N$
• D. $141.4N$

Answer 1

Answer: (B)

$98.9N$

Initial height of the ball $h_i=2.5m$

Height upto which the ball raised after collision $h_f=0.4m$

Thus coefficient of restitution $e^2=\frac{h_f}{h_i}=\frac{0.4}{2.5}$

$⟹e=0.4$

Velocity with which the ball hits the ground $v_i=\sqrt{2g{h}_i}=\sqrt{2\times 9.8\times 2.5}=7m/s$ (downwards)

Velocity of the ball just after collision in upward direction $v_f=e{v}_i=0.4\times 7=2.8m/s$ (upwards)

Mass of ball $m=200g=0.2kg$

Change in momentum $\Delta P=m{v}_f-m(-v_i)=m(v_f+v_i)=0.2\times (7+2.8)=1.96m/s$

Time of contact $\Delta t=0.02s$

Force exerted $F=\frac{\Delta P}{\Delta t}=\frac{1.96}{0.02}=98N$