IIT JEE , AIEEE Forum - Physics , Mechanics

iitaspirant10

A block of mass 200g is suspended through a vertical spring.The spring is stretched by 1.0cm when the block is in equilibrium.A particle of mass 120g is dropped on the block from a height of 45cm.The particle sticks to the block after the impact.Find the maximum extension in the spring.[g=10m/s²]

skb

is it 4.5cm?

ABHI_23

IS THE ANS 6cm approx

ganesha1991

kx=mg
k = 0.2*10/10^-2 = 200
mgh=1/2mv^s
v = root 2*g*.45
= 3rootg
120(3rootg) + 200(0)=320(V)
V = 9rootg/8
now -1/2mv^s=1/2kx^2 -mgx

ganesha1991

solve the quadratic equation and get the answer

skb

ganpat ji,u did calculation mistake in finding v.
v=root.2*10*.45
=3 m/s.
iitaspirant wats the ans?

varun.tinkle

THIS SUM IS LITTLE TRICKY
FIRST WE CALCULATE THE VELCOITY OF THE MASS 100 g AFTER FALLING =3MS
CONSERVING MOMENTUM WE GET THAT THE VELOCITY OF THE WHOLE SYSTEM IS 9/8
NOW LET THE SPRING COMPRESS BY A DISTANCE X
SO
1/2(M+m)(9/8)^2+MGX=1/2KX^2
SOLVE AND GET THE ANSWER
PLS RATE ME IF U FIN ME USEFUL
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ABHI_23

yes i have also done like varun then ans= 6cm approx

iitaspirant10

Answer is 6.1cm n am not getting it :(

iitaspirant10

wait, thats is precisely what i did...mustve made an error somwer

ABHI_23

Should i give you a solution or not????

iitaspirant10

sure..go ahead

ABHI_23

HERE it is given that string is already stretched

so we can find value of K so K= 0.2*10/0.01

therefore k= 200N now it is at ht 45cm so vel when it hits block so

vel at that pt = = 3m/s and this results in plastic colln with block

so by conservation of momentum 0.12*3 = (0.12+0.2) V SO V = 1.125m/s

now by conservation of energy

so by putting k = 200N , v= 3m/s , V= 1.125m/s

m = 0.12 M= 0.2

we get x = 6cm and if we take g = 9.8m/s^2

we get it as 6.1cm approx

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