Question

A ball of mass = 100 gm is released from a height h1 = 2.5 m from the ground level and then rebounds
to a height h2 = 0.625 m. The time of contact of the ball and the ground is At = 0.01 sec. The impulsive
(impact) force offered by the ball on the ground is :
(A) 105 N
(B) 1.05 N
(C) 2.08 N
(D) 208 N​

Answer 1

ANSWER:

The impulsive force offered by the ball on the ground is 105 N.

EXPLANATION:

The impulsive force is the force exerted on the ground when any object is having an impact or hit the ground.

Impulsive force= (Mass of the object which impacted on ground*(difference in velocities at different heights))/(time period)

Mass of the object is 100 gm and change in velocity of the ball at different positions can be determined as  

$\left|v_{2}-v_{1}\right|=v_{2}+v_{1}As$

$v_{i}=\sqrt{2 g h_{i}}|v_2-v_1 |=v_2+v_1=√2g(√(h_1 )+√(h_2 ))$

Thus,

$F=\frac{100}{1000}k g * \frac{√({2 * 9.8} )}{0.01}*(√({0.625} )+√({2.5} ))\frac{m s^{-1}  }{s}$

$F=0.1 \mathrm{kg}* \frac{√({19.6} )}{0.01}*(0.791+1.581)\frac{\mathrm{ms}^{-1}  }{\mathrm{s} }$

$F=10 \mathrm{kg}* 4.427 * 2.372 \frac{\mathrm{ms}^{-1}  }{\mathrm{s} }$

F=105 N

Thus, the impulsive force offered by the ball on the ground is 105 N.