A ball of mass 4 kg, moving with a velocity of 10 ms^-1collides with a spring of length 8m and force constant 100 Nm^-1. The length of the compressed spring is xm . The value of X , to the nearest in…

Question

A ball of mass 4 kg, moving with a velocity of 10 ms^-1collides with a spring of length 8m and force constant 100 Nm^-1. The length of the compressed spring is xm . The value of X , to the nearest integer, is \_ \_ \_ \_

Accepted Answer

Correct answer: 6. (6) JEE Main 2021 Let's say the compression in the spring by: y. So, by work energy theorem we have ⇒ 12 mv^2= 12 ky^2 ⇒y= mk v ⇒y= 4100 10 ⇒y=2m ⇒final length of spring =8-2=6m

$A ball of mass 4 kg, moving with a velocity of 10 ms^{- 1} collides with a spring of length 8 m and force constant 100 Nm^{- 1} . The length of the compressed spring is xm . The value of X, to the nearest integer, is____$

Solution

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