JEE MAIN - Physics (2021 - 18th March Evening Shift - No. 21)
A ball of mass 4 kg, moving with a velocity of 10 ms$$-$$1, collides with a spring of length 8 m and force constant 100 Nm$$-$$1. The length of the compressed spring is x m. The value of x, to the nearest integer, is ____________.
Answer
6
Explanation
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If spring compressed by x,
then work done by spring = 0 $$-$$ $${1 \over 2}$$ $$\times$$ 4 $$\times$$ 102
Applying work energy theorem,
$$-$$$${1 \over 2}$$ kx2 = $$-$$$${1 \over 2}$$ $$\times$$ 4 $$\times$$ 102
$$ \Rightarrow $$ 100x2 = 4 $$\times$$ 102
$$ \Rightarrow $$ x = 2
$$ \therefore $$ Final length of the spring = 8 $$-$$ 2 = 6 m
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