A ball of mass 2kg is dropped from a height of 0.4m above the free end of a vertically fixed spring of constant $k = 200 N m^{- 1}$

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The maximum K.E is attained 0.5 m below original position P.

The maximum potential energy attained by the spring is 9 J

The maximum K.E attained by the ball is 9 J

The maximum potential energy attained by the spring is 16 J

answer is A, B, C.

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Detailed Solution

The ball moves with acceleration, till the neutral position, where $m g = k x$
$\Rightarrow 2 \times 10 = 200 \times x \Rightarrow x = 0.1 m .$ This is the maximum
Velocity (KE) position.
(b) is correct
Energy equation at maximum KE:
$\frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = m g (h + x)$
$\Rightarrow K E_{\max} + \frac{1}{2} \times 200 \times {(0.1)}^{2} = 2 \times 10 (0.5)$
$\Rightarrow K E_{\max} = 10 - 1 = 9 J$
(a) is correct
Energy equation at maximum spring energy (maximum compression)
$m g (h + x_{m}) = \frac{1}{2} k x_{m^{2}}$
$\Rightarrow 20 (0.4 + x_{m}) = \frac{1}{2} \times 200 {(x_{m})}^{2}$
$\Rightarrow 100 x_{m^{2}} - 20 x_{m} - 8 = 0$
Energy equation at maximum spring energy (maximum compression)
$m g (h + x_{m}) = \frac{1}{2} k x_{m^{2}}$

$\Rightarrow 20 (0.4 + x_{m}) = \frac{1}{2} \times 200 {(x_{m})}^{2}$
$\Rightarrow 100 x_{m^{2}} - 20 x_{m} - 8 = 0$
$\Rightarrow 25 x_{m^{2}} - 5 x_{m} - 2 = 0$
$x_{m} = \frac{5 \pm \sqrt{25 + 200}}{50} = \frac{20}{50} = 0.4 m$
$P E_{\max} = \frac{1}{2} \times 200 \times 0.16 = 16 J$
(C) is correct
$∴ (a), (b) and (c)$ are correct