Physics, asked by SparklingBoy, 1 day ago

A 1 kg block moves towards a light Spring with velocity of 8 m/s. When the spring compressed by 6m its Momentum becomes one fourth to the original Momentum.
Find the spring constant of The spring,

❒ Note : Try To Use LaTex ​

Answers

Answered by NewGeneEinstein
15

\underline{\underline{\Large{\bf Given:-}}}

\bull\sf Mass\:of\:block=m=1kg

\bull\sf Velocity=v_1=8m/s

\bull\sf Compression=x=6m

\underline{\underline{\Large{\bf To\:find-}}}

\diamond\sf Spring\:constant=k

\underline{\underline{\Large{\bf Solution:-}}}

Let

\bull\sf initial\:momentum=P_1

\bull\sf Final\:momentum=P_2

\bull\sf Velocity=V_2

We know

\boxed{\sf Momentum_{(P)}=Mass_{(m)}\times Velocity_{(v)}}

\\ \sf\longmapsto P_1=mv_1

\\ \sf\longmapsto P_1=1\times 8

\\ \bf\longmapsto P_1=8kgm/s

ATQ

\\ \sf\longmapsto P_2=\dfrac{P_1}{4}\dots(1)

\\ \sf\longmapsto P_2=mv_2\dots(2)

From both equations

\\ \sf\longmapsto \dfrac{P_1}{4}=mv_2

\\ \sf\longmapsto \dfrac{8}{4}=1v_2

\\ \sf\longmapsto 2=1v_2

\\ \sf\longmapsto v_2=2m/s

Now

Consumed mechanical energy:-

\boxed{\sf \dfrac{1}{2}mv_1^2+0=\dfrac{1}{2}mv_2^2+\dfrac{1}{2}kx^2}

\\ \sf\longmapsto \dfrac{1}{2}\left(mv_1^2=mv_2^2+kx^2\right)

\\ \sf\longmapsto mv_1^2=mv_2^2+kx^2

Putting values

\\ \sf\longmapsto 1(8)^2+1(2)^2+k(6)^2

\\ \sf\longmapsto 64=4+36k

\\ \sf\longmapsto 36k=64-4

\\ \sf\longmapsto 36k=60

\\ \sf\longmapsto k=\dfrac{60}{36}

\\ \sf\longmapsto k=\dfrac{5}{3}N/m

\\ \therefore{\underline{\boxed{\bf k=\dfrac{5}{3}N/m}}}

Answered by Atlas99
18

GIVEN THAT:-

Mass = 1kg

Velocity = 8m/s

Compression = 6m

TO FIND:-

The Spring Constant

__________________________________________

Here,

Mass is m

Velocity is V1

Compression is x

Spring Constant

___________________________________________

SOLUTION:-

Let initial momentum,final momentum and velocity be P1,P2 and V2 respectively.

P1 = mv1 = 1 × 8 = 8kgm/s

P2 = mv2

P1/4 = 1×V2

8/4 = V2

V2 = 2m/s.

Consumed mechanical energy

1/2 mv^2 + 0 = 1/2mv^2 + 1/2 kx^2

1 2

1(8)^2 + 1(2)^2 + k(6)^2

64 = 4+36k

36k = 64 - 4

36k = 60

k = 60/36

k = 5/3 N/m

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