Question

A 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 ​

Answer 1

Info:- You missed 1kg in question .

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

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

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

$\bull\sf Compression=x=6m$

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

$\diamond\sf Spring\:constant=k$

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

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

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

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

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

$\\ \sf\longmapsto 36k=60$

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

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

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

Answer 2

★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 is k

★SOLUTION:

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

= 8kgm/s P1 = mv1 = 1x 8 = P2 = mv2

P1/4 = 1xV2

8/4 = V2

V2 = 2m/s.

Consumed mechanical energy(please view attachment)

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