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One end of a light elastic string of natural length $0.5{\text{ }}m$ and modulus of elasticity $12{\text{ }}N$ is attached to a fixed point $O$. The other end of the string is attached to a particle $P$ of mass $0.24{\text{ }}kg$. $P$ is projected vertically upwards with speed $3{\text{ }}m{\text{ }}{s^{ - 1}}$ from a position $0.8{\text{ }}m$ vertically below $O$.

a) Calculate the speed of the particle when it is moving upwards with zero acceleration.

b) Show that the particle moves 0.6 m while it is moving upwards with constant acceleration.

پاسخ تشریحی :
نمایش پاسخ

a) $0.24g = 12\left( x \right)/0.5$

$x = 0.1$

EITHER

$\frac{1}{2} \times {\text{0}}.24 \times {3^2} + 12 \times {\left( {0.8 - 0.5} \right)^2}/\left( {2 \times 0.5} \right) = $

$0.24{v^2}/2 + 12 \times {0.1^2}/\left( {2 \times 0.5} \right)$

$ + 0.24g\left( {0.8 - 0.5 - 0.1} \right)$

$v = 3.61{\text{ }}m{s^{ - 1}}$

OR

$0.24vdv/dx = mg - 12x/0.5$

$0.24{v^2}/2 = 2.4x - 12{x^2}{\text{ }}\left( { + c} \right)$

$v = 3,{\text{ }}x = 0.3,{\text{ }}c = 1.44$

$x = 0.1,{\text{ }}v = 3.61{\text{ }}m{s^{ - 1}}$

b) $0.24 \times {3^2}/2 + 12 \times {\left( {0.8 - 0.5} \right)^2}/\left( {2 \times 0.5} \right) = $

$0.24g\left( {0.8 + x} \right)$

$x = 0.1m$

$s = \left( {0.5 + 0.1} \right) = 0.6{\text{ }}m$

OR

$\frac{1}{2} \times 12 \times {0.3^2}/0.5 + \frac{1}{2} \times 0.24 \times {3^2}$

$ = \frac{1}{2} \times 0.24{v^2} + 0.24 \times 10 \times 0.3$

$v = \sqrt {12} $

Either

$0 = 12 - 2 \times 10s$

$s = 0.6$

OR $\frac{1}{2} \times 0.24 \times 12 = 0.24 \times 10s$

$s = 0.6$

OR

$\frac{1}{2} \times 12 \times {0.3^2}/0.5 + \frac{1}{2} \times 0.24 \times {3^2}$

$ = 0.24 \times 10y$

$y = 0.9$

$s = 0.9 - 0.3 = 0.6$

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