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)#

2 Answers
Ultrilliam
May 27, 2018

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

Explanation:

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

Because:

  • #v_o lt sqrt(2gl_o)#

Then:

  • #1/2m v_o^2 lt m g l_o #

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 #h# above its starting point, then:

#1/2 m v_o^2 = mgh implies h = v_o^2/(2g) #

A08
May 27, 2018

(5)

Explanation:

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

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

Inserting various values we get

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

It is given that

#v_0" < "sqrt(2l_0g)#
#=>"Initial KE"=1/2mv_0^2" < "1/2mxx2l_0g#
#=>"Initial KE < "mgl_0#
#=>"Initial KE"# is not sufficient to raise the ball from the mean position #y=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)#

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