Question

A ball is dropped from a height of 5 m onto a sandy floor and penetrates the sand up to 10 cm before coming to rest. Find the retardation of the ball is sand assuming it to be uniform.

Answer 1

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Answer 2

The retardation of the ball is sand assuming it to be uniform is 500 m/s².

Explanation:

Step 1:

Given in the questions

Initial ball velocity (u) is 0 m / s.

Let the ball's final velocity  when it's just entering the sand is v m / s.

Acceleration (a)  = g = 9.8 m/s² ≈ 10 m/s²

A ball falls from 5 m in height

                 s = 5 m

Step 2:

Applying the laws of Newton,  

we get,

           $v^{2}-u^{2}=2 a s$

           $v^{2}-0^{2}=2 \times 10 \times 5$

            $v^{2}=20 \times 5$

            $v^{2}=100$

            $v=\sqrt{100}$  

            $v=10 \frac{m}{s}$

Step 3:

Now, the initial velocity (v) is 10 m / s when the ball hits the sand

And the final velocity v ' is 0 m / s if the ball stops inside the sand

There is a 10 cm or 0.1 m displacement inside the sand

And the ball is accelerated inside the sand by a' m /s².

Again, applying the laws of Newton, we get,

                                  $v^{\prime 2}-v^{2}=2 a^{\prime} s$

                                 $0^{2}-10^{2}=2(-a) \times 0.1$

                                         $100=-2 a \times 0.1$

                                         $100=-0.2 a$

(The negative sign here indicates ball retardation, i.e.negative acc..)

                                         $0.2 a=100$

                                              $a=\frac{100}{0.2}$

                                              $a=500 \mathrm{m} / \mathrm{s}^{2}$

Thus, the ball retardation is 500 m /s².