Question

A steel ball of mass $50g$ is thrown vertically downwards with a velocity of $15m{s}^{-1}$ from a height of $20m$. It buries itself in the sandy ground to a depth of $20cm$. The magnitude of average force exerted on the ball by the sand is nearly: $(g=10m{s}^{-2}$, Assume resistance offered by sand is constant throughout)

• A. $49$ N
• B. $78.6$N
• C. $98$ N
• D. $29.4$ N

Answer 1

The correct option is B $78.6$N

Using the equation of motion: $v^2=u^2+2as$
$v^2=225+2\times 10\times 20=225+400=625\Rightarrow v=25m/s$

Again using the third equation of motion, when ball finally comes to rest in the sandy ground after covering distance of 20 cm: (Let $\alpha$ be the deceleration for this phase of motion in sandy ground)

$0^2={25}^2-2\times \alpha \times 0.20=625-0.4\alpha$

$\alpha =-1562.5m/s^2$

The forces acting on the ball are $mg$ and the force exerted by the sand $F$.
$\therefore$ Deceleration, $d=\frac{F-mg}{m}$
$1562.5=(F-0.5)/\left(0.05\right)$

$F=78.125+0.5=78.625N$