Question

A spherical ball of mass 10-6 kg hits a wall 1000 times per second normally with a

velocity of 1000 m/s and rebounds with same velocity along the initial direction.

The force experienced by the wall is

(a) 1 N (b) 4 N (c) 2 N (d) 8 N​

Answer 1

Given:

A spherical ball of mass 10-6 kg hits a wall 1000 times per second normally with a velocity of 1000 m/s and rebounds with same velocity along the initial direction.

To find:

Force experienced by wall ?

Calculation:

Force experienced by the wall can be described as the change in net momentum (transferred to wall) divided by the total time taken.

$\Delta P_{ball} = m\Delta v$

$\implies \Delta P_{ball} = {10}^{ - 6} \bigg( - 1000 - 1000 \bigg)$

$\implies \Delta P_{ball} = {10}^{ - 6} \times \bigg( - 2000 \bigg)$

$\implies \Delta P_{ball} = - 2 \times {10}^{ - 3} \: kgm {s}^{ - 1}$

Momentum change transferred to wall :

$\implies \Delta P_{wall} = - \Delta P_{ball} = 2 \times {10}^{ - 3} \: kgm {s}^{ - 1}$

Net Momentum transferred to wall :

$\implies \Delta P_{net} = 2 \times {10}^{ - 3} \times (no. \: of \: times \: hitting)$

$\implies \Delta P_{net} = 2 \times {10}^{ - 3} \times {10}^{3}$

$\implies \Delta P_{net} = 2 \: kgm {s}^{ - 1}$

So, force experienced by wall:

$\therefore \: F = \dfrac{ \Delta P_{net} }{dt}$

$\implies \: F = \dfrac{2}{1}$

$\implies \: F = 2 \: N$

So, force experienced by wall is 2 N.

Answer 2

Explanation:

$(c)2 \: n$

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