Question

A 1 kg block move towards a light spring with a velocity 8m/s.
When the spring is compressed by 3m, its momentum
becomes half of the original momentum. Spring constant of
the spring is:
(A) 18/3 N/m
(B) 16/3 N/m
(C) 3 N/m
(D) 8 N/m​

Answer 1

Answer:

answer is b

Explanation:

becomes half of the original momentum

Answer 2

Spring constant of  the spring is $\dfrac{16}{3}\ N\ m^{-1}$ .

It is given that momentum of the body becomes half the initial .

Therefore , its velocity after compression is 4 m/s .

We know , total energy is conserved in SHM .

So , initial energy = final energy

$\dfrac{mv^2}{2}=\dfrac{kx^2}{2}+\dfrac{m(\dfrac{v}{2})^2}{2}\\\\k=\dfrac{3mv^2}{4x^2}$

Putting all values in above equation .

We get :

$k=\dfrac{16}{3}\ N\ m^{-1}$

Learn More :

Simple Harmonic Motion

https://brainly.in/question/12599713