Question

a block c of mass m is moving with velocity Vo and collides elastically with block A of mass m and connected to another block B of mass 2 m through spring contant K. what is K ? if X0 is the compression of spring when velocity of A and B is same

Answer 1

your question is -> A block C of mass m is moving with velocity $v_0$ and collides elastically with block A of mass m and connected to another block B of mass 2m through spring of spring constant k. What is k, if $x_0$ is compression of spring when velocity of A and B is same?

solution : let velocity of block of mass A and block c is v.

from law of conservation of linear momentum,

initial momentum = final momentum

⇒ $mv_0=mv+2mv$

⇒$mv_0=3mv$

⇒$v=\frac{v_0}{3}$

Using conservation of energy,

kinetic energy of block C = kinetic energy of system of block A and B + spring potential energy

⇒$\frac{1}{2}mv_0^2=\frac{1}{2}3mv^2+\frac{1}{2}kx_0^2$

⇒$mv_0^2=3m\left(\frac{v_0}{3}\right)^2+kx_0^2$

⇒$\frac{2}{3}mv_0^2=kx_0^2$

⇒k = $\frac{2mv_0^2}{3x_0^2}$

Answer 2

Concept :-->

** when she strikes a acquire the velocity of c and C becomes stationary.

A force B to move and ultimately both Aquire common velocity with spring compressed by x°

**

here

V = mv°/m + 2m

V = v°/3

according to law of conservation of mechanical energy

1/2m(v°)² = 1/2k(x°)² + 1/2(3m)[v°/3]²

cancelling 1/2 m from both sides.

on solving

k = (2/3 )* m(v°)²/(x°)²

Khushi here✔️✌