If a ball of mass 0.1 kg hits the ground from the height of 20 m and bounce back to the same height then find out the force exerted on the ball if the time of impact is 0.04 sec. (g = 10 m/s2)

Question

If a ball of mass 0.1 kg hits the ground from the height of 20 m and bounce back to the same height then find out the force exerted on the ball if the time of impact is 0.04 sec. (g = 10 m/s2) (A) -100N (B) -10N (C) -1N (D) None of these

Tardigrade Admin · Accepted Answer

The velocity the ball before hitting the ground,

v_{1}^{2} = u^{2} + 2 g h

u = 0

so,

v_{1} = 2 g h = 2 \times 10 \times 20

= 20m/s in downward direction
After hitting the ground, ball reaches to the same height. So the collision is elastic. So velocity just after hitting the ground v₂= 20 m/s in upward direction
From impulse - momentum theorem,
Applied force = Rate of change of momentum
Force

= \frac{Δ P}{Δ t} = \frac{m v _{2} - m v _{1}}{Δ t}

= \frac{m ( v _{2} - v _{1} )}{Δ t} = \frac{- 0.1 \times 40}{0.04} = - 100 N

[in upward direction]

Q. If a ball of mass 0.1kg hits the ground from the height of 20m and bounce back to the same height then find out the force exerted on the ball if the time of impact is 0.04sec. (g=10m/s2)

-100N

-10N

-1N

None of these

Q. If a ball of mass $0.1 k g$ hits the ground from the height of $20 m$ and bounce back to the same height then find out the force exerted on the ball if the time of impact is $0.04 sec$ .
$(g = 10 m / s^{2}$ )