JEE MAIN - Physics (2022 - 26th July Evening Shift - No. 3)
A ball of mass $$0.15 \mathrm{~kg}$$ hits the wall with its initial speed of $$12 \mathrm{~ms}^{-1}$$ and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is $$100 \mathrm{~N}$$, calculate the time duration of the contact of ball with the wall.
0.018 s
0.036 s
0.009 s
0.072 s
Explanation
F = 100 N
$$\Delta$$P = 2 $$\times$$ 0.15 $$\times$$ 12
= 3.6
$$\Rightarrow$$ t = $${{3.6} \over {100}}$$ = 0.036 s
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