Question

A spring with a constant of `6/5 (kg)/s^2` is lying on the ground with one end attached to a wall. An object with a mass of `3/5 kg` and speed of `5/4 m/s` collides with and compresses the spring until it stops moving. How much will the spring compress?

Answer 1

`(5 sqrt(2))/8 "m"`

Explanation:

Kinetic energy of object = Energy stored in spring

`1/2mv^2 = 1/2kx^2`

`x = vsqrt(m/k) = 5/4 "m/s"sqrt((3/5"kg")/(6/5 "kg/s"^2)) = 5/(4sqrt(2)) "m" = (5sqrt(2))/8 "m"`

Answer 2

The compression is `=0.88m`

Explanation:

Mass of the object is `m=3/5kg`

Speed of the object is `v=5/4ms^-1`

The kinetic energy of the object is

`KE=1/2mv^2=1/2*3/5*(5/4)^2=0.46875J`

This energy will be stored in the spring

`E=1/2kx^2`

The spring constant is `k=6/5kgs^-2`

Let the compression of the spring be `=xm`

Therefore,

`0.46875=1/2*6/5*x^2`

`x^2=(5*0.46875)/3=0.78m^2`

The compression of the spring is

`x=sqrt0.78=0.88m`