Average Force Exerted by Sand on Penetrating Ball

To find the average force exerted by the sand on the ball, we can apply the Work-Energy Theorem or use Kinematics with Newton's Laws of Motion.

Method 1: Work-Energy Theorem

Let the motion be analyzed from the initial release point (height $h$) to the final point where the ball comes to rest inside the sand (depth $d$).

• Given data:

  • Mass of the ball, $m = 2\text{ kg}$
  • Height from which the ball falls, $h = 10\text{ m}$
  • Depth of penetration into the sand, $d = 10\text{ cm} = 0.1\text{ m}$
  • Acceleration due to gravity, $g = 10\text{ m/s}^2$

• Work done by forces:

  1. Work done by gravity ($W_g$): Gravity acts downwards over the total vertical displacement $(h + d)$. $W_g = mg(h + d)$

  2. Work done by the sand ($W_{\text{sand}}$): The average force exerted by the sand ($F_{\text{sand}}$) acts upwards, opposing the motion through distance $d$. $W_{\text{sand}} = -F_{\text{sand}} \cdot d$

According to the Work-Energy Theorem, the net work done on the ball is equal to the change in its kinetic energy ($\Delta K$): $W_{\text{net}} = \Delta K$

Since the ball starts from rest ($K_i = 0$) and comes to rest at the end ($K_f = 0$): $W_g + W_{\text{sand}} = 0$

$mg(h + d) - F_{\text{sand}} \cdot d = 0$

Rearranging for $F_{\text{sand}}$: $F_{\text{sand}} = mg \left( \frac{h + d}{d} \right)$

Substitute the given values: $F_{\text{sand}} = (2)(10) \left( \frac{10 + 0.1}{0.1} \right)$

$F_{\text{sand}} = 20 \times \frac{10.1}{0.1} = 20 \times 101 = 2020\text{ N}$

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Method 2: Kinematics and Newton's Second Law

1. Velocity just before hitting the sand ($v$): $v^2 = u^2 + 2gh = 0 + 2(10)(10) = 200\text{ m}^2/\text{s}^2$

2. Deceleration inside the sand ($a$): Using $v_f^2 = v^2 - 2ad$ with final velocity $v_f = 0$: $0 = 200 - 2a(0.1)$ $0.2a = 200 \implies a = 1000\text{ m/s}^2$

3. Applying Newton's Second Law: The net force acting upwards on the ball inside the sand is: $F_{\text{net}} = ma = 2 \times 1000 = 2000\text{ N}$

   Since $F_{\text{net}} = F_{\text{sand}} - mg$: $F_{\text{sand}} = F_{\text{net}} + mg = 2000 + (2 \times 10) = 2020\text{ N}$

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Thus, the average force exerted by the sand on the ball is $2020\text{ N}$.