Question

A mass of 2 kg is moving with a speed of 2m/s on a horizontal smooth surfaceand collides with a massless of spring constant k=200n/m. The maximum compression on the spring will be

Answer 1

Plz refer to the attachment

As we know, by the conservation of energy

Decrease in kinetic energy= increase in potential energy(or spring force)

$\frac{1}{2} m {v}^{2} = \frac{1}{2} k {x}^{2} \\ \frac{1}{2} .2. { 2}^{2} = \frac{1}{2} .200. {x}^{2} \\ on \: solving \\ x = \frac{2}{10} = 0.2 \:m$

I have taken m= 2 kg, v= 2 m/s, k= 200 N/m and x be the the compression of the string

Answer 2

Answer:

The maximum compression will be 0.015m

Explanation:

Compression is volume of size of any object which is reduced by applying any stress on that specific object. For this use the formula:

$x_{0} =\sqrt{\frac{m}{k}$  X v

Given:

Mass objects is 2Kg speed is 2m/s and spring is 200 n.m

Find: Need to find out the compression highest value on the object

Solution:

Need to put the value in the formula:

 $x_{0} =\sqrt{\frac{m}{k}$ X v=$x_{0} =\sqrt{\frac{2}{200}$X 1.5

= 0.01 x 1.5

=0.015

So, the final answer is 0.015

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