You visited us 1 times! Enjoying our articles? Unlock Full Access!

Com of Solid Bodies

A spring is h...

Question

A spring is held between two blocks of masses $m_{1}$ and $m_{2}$ but not connected to them and then the blocks are moved towards each other so as to compress the spring. If the blocks are suddenly released, and if $K_{1}$ and $K_{2}$ are the kinetic energies of the blocks when the spring just gets detached from the blocks, then the ratio $K_{1} : K_{2}$ must be:

m_{1} : m_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

m_{2} : m_{1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

m_{1}^{2} : m_{2}^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

m_{2}^{2} : m_{1}^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $m_{2} : m_{1}$

Using linear momentum conservation:
$m_{1} v_{1} = m_{2} v_{2}$
$\Rightarrow v_{2} = \frac{m_{1}}{m_{2}} v_{1}$
$K . E_{1} b l o c k = \frac{1}{2} m_{1} v_{1}^{2}$
$K . E_{2} b l o c k = \frac{1}{2} m_{2} v_{2}^{2}$ $= \frac{1}{2} m_{2} \frac{m_{1}^{2}}{m_{2}^{2}} v_{1}^{2}$

$K . E_{1} : K . E_{2} = m_{2} : m_{1}$

Suggest Corrections

Similar questions

m_{1}

m_{2}

m_{1}

m_{2}

d

m_{1}

m_{1}

m_{2}

Two bodies of masses m₁ and m₂ having equal kinetic energies. If p₁ and p₂ are their momentum respectively, Then the ratio of p₁ and p₂ is.

M_{1}

M_{2}

K

t = 0

M_{1}

M_{2}

W_{1}

W_{2}

\frac{W_{1}}{W_{2}}

m_{1}

m_{2}

k

m_{2}

v_{o}

Join BYJU'S Learning Program