JEE MAIN - Physics (2022 - 27th July Evening Shift - No. 5)
The velocity of the bullet becomes one third after it penetrates 4 cm in a wooden block. Assuming that bullet is facing a constant resistance during its motion in the block. The bullet stops completely after travelling at (4 + x) cm inside the block. The value of x is :
2.0
1.0
0.5
1.5
Explanation
S = 4 cm
$$v{'_4} = {v \over 3}$$, a = constant
$${v_{4 + x}} = 0$$
$$\left( {{v^2} - {{{v^2}} \over a}} \right) = 2a(4)$$
$$({v^2} - 0) = 2a(4 + x)$$
$${4 \over {4 + x}} = {8 \over 9}$$
$$ \Rightarrow x = 0.5$$ m
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