Question

A block of mass m sliding on a smooth horizontal surface with a velocity v meets a long horizontal spring fixed at one end and having spring constant k as shown in the figure. Find the maximum compression of the spring. Will the velocity of the block be the same as v when it comes back to the original position shown?
Concept of Physics - 1 , HC VERMA , Chapter " Work and Energy"

Answer 1

Assume that the spring compression is x .
Here , the Initial kinetic energy of the block 
K.E = 1/2 mv²

When the spring is compressed at it maximum range at that point the Kinetic energy will be changed to potential energy .
P.E = 1/2 mx²

Now from the law of conservation of energy .we get
1/2 mv² = 1/2 kx²
mv² = kx²
v² = kx²/m
x² = (v² × m) /k
The maximum compression x, is $x = v \sqrt{ \frac{m}{k} }$

⇒Now for the second case , we assume that there is no loss in the spring. So the kinetic energy is the same when spring comes to it's original position.
But velocity is a vector quantity , so it will not be same as like original . 


Hope it Helps.

Answer 2

Answer:

This can be solved more easily by the energy conservation law:

K.E.i + P.E.i = K.Ef + P.E.f

Let compression in the spring be 'x'

initially

velocity= v

compression in spring(x)= 0

finally

velocity= 0

compression in spring= 'x'

substituting values we get.....

½·m·v² + ½·k·(0)² = ½·m·(0)² + ½·k·x²

therefore,

½mv² = ½kx² .....(i)

x = v√(m/k)

Now...... initial velocity of the block

(v) = x√(k/m) [towards the spring]

but velocity of the block after the spring pushes it to regain its original position i.e.(x=0) would be same by magnitude but in direction opposite to the spring .......hence the velocity will not be the same.....

hope this helps