Question

A block sliding along a horizontal frictionless surface with speed v collides with a spring and compresses it by 1.0 cm . What will be the compression if the same block collides with the spring at a speed of 4v

Answer 1

The compression is
`x = 0.05 \ m`

Step-by-step explanation:

From the question we are told that

The first velocity is v

The compression of the spring is
`d = 1.0 \ cm = 0.01 \ m`

The second velocity is 4v

Generally according to the law of energy conservation

The kinetic energy of the block is equal to the energy stored in the spring that is

`(1)/(2) * m* v^ 2 = (1)/(2) * k * d^2`

For first speed

`m* v^ 2 = k * 0.01^2`

=>
`m* v^ 2 = k * 0.0001`

=>
`k = ( m v^2 )/( 0.0001)`

For second speed

`(1)/(2) * m* (5v)^ 2 = (1)/(2) * k * x^2`

=>
`12.5mv^2 = 0.5 k x^2`

substituting for k

=>
`12.5mv^2 = 0.5 ((mv^2)/(0.0001) ) x^2`

=>
`12.5 = 5000x^2`

=>
`x = 0.05 \ m`