How to find the maximum distance traveled by a ball of mass M, tied to a light elastic string, projected vertically downwards with an initial velocity of Vo?

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It is given that;
spring constant -k
Initial length - Lo
Vo < sqrt(2gLo)2gLo

2 Answers
May 27, 2018

h = v_o^2/(2g) h=v2o2g

Explanation:

Due to conservation of energy, the particle will pass via its starting point with upward velocity v_ovo.

Because:

  • v_o lt sqrt(2gl_o)vo<2glo

Then:

  • 1/2m v_o^2 lt m g l_o 12mv2o<mglo

ie the particle will stop before the string extends. All of its energy will be in the form of gravitational PE.

If it rises to height hh above its starting point, then:

1/2 m v_o^2 = mgh implies h = v_o^2/(2g) 12mv2o=mghh=v2o2g

May 27, 2018

(5)

Explanation:

The ball rises from the lowest point crosses y=0y=0 and moves upwards. At the highest point it comes to rest momentarily. Let that position be =y=y. Using Law of Conservation of energy we get

"Mechanical PE of string"+"GPE"="Initial KE"Mechanical PE of string+GPE=Initial KE

Inserting various values we get

"Mechanical PE of string"+mgy=1/2mv_0^2Mechanical PE of string+mgy=12mv20 ......(1)

It is given that

v_0" < "sqrt(2l_0g)v0 < 2l0g
=>"Initial KE"=1/2mv_0^2" < "1/2mxx2l_0gInitial KE=12mv20 < 12m×2l0g
=>"Initial KE < "mgl_0Initial KE < mgl0
=>"Initial KE"Initial KE is not sufficient to raise the ball from the mean position y=0y=0 to its unstreched length l_0 on its way up.

As the string remains slack, its mechanical PE=0. :.Equation (1) reduces to

mgy=1/2mv_0^2
=>y=v_0^2/(2g)