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 ____________. Numerical answer type. Work it out, then open the answer.

Answer: 6. 1/2mv² = 1/2kc², where c is the compression. c = v√(m/k) = 10√(4/100) = 2 m Compressed length x = 8 - 2 = 6 m So x = 6.

A ball of mass 4 kg, moving with a velocity of 10 ms ⁻¹, collides… | JEE Main 2021 (18 Mar, Shift 2)

Q15JEE Main 2021 · 18 MAR · SHIFT 2Easy

Spring & Elastic Energy

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 ____________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 6

$\frac{1}{2}mv^2 = \frac{1}{2}kc^2$ , where $c$ is the compression.
$c = v\sqrt{\frac{m}{k}} = 10\sqrt{\frac{4}{100}} = 2$ m
Compressed length $x = 8 - 2 = 6$ m
So $x = 6$ .

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Spring & Elastic Energy question from JEE Main 2021 (18 Mar, Shift 2)

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