A block ' <math> <mi>A</mi> </math> ' of mass <math> <mi>m</mi> </math> attached to a massless spring is performing oscillatory motion of amplitude ' <math> <mi>A</mi> </math> ' on a frictionless horizontal plane. If it collides elastically with another block ' <math> <mi mathvariant="normal">B</mi> </math> ' of mass <math> <mn>2</mn> <mtext> </mtext> <mi mathvariant="normal">m</mi> </math> kept initially at rest, when ' <math> <mi mathvariant="normal">A</mi> </math> ' is passing through its equilibrium point, the amplitude of oscillation for the block ' <math> <mi mathvariant="normal">A</mi> </math> ' become <math> <mi>f</mi> </math> A. The value of <math> <mi>f</mi> </math> is:

v m a x = A ω = A k m Momentum conservation m v max + 0 = m v A + 2 m v B v A + 2 v B = v m a x v 2 - v 1 = u 1 - u 2 ⇒ v B - v A = v m a x By (1) & (2) v A = - A ω 3 v A = A new ω A ω 3 = A new ω ∴ A new = A 3 ; f = 1 3

A block ' <math> <mi>A</mi> </math> ' of mass <math> <mi>m</mi> </math> attached to a massless spring is performing oscillatory motion of amplitude ' <math> <mi>A</mi> </math> ' on a frictionless horizontal plane. If it collides elastically with another block ' <math> <mi mathvariant="normal">B</mi> </math> ' of mass <math> <mn>2</mn> <mtext> </mtext> <mi mathvariant="normal">m</mi> </math> kept initially at rest, when ' <math> <mi mathvariant="normal">A</mi> </math> ' is passing through its equilibrium point, the amplitude of oscillation for the block ' <math> <mi mathvariant="normal">A</mi> </math> ' become <math> <mi>f</mi> </math> A. The value of <math> <mi>f</mi> </math> &nbsp;is: &lt;!-- [if !supportLineBreakNewLine]--&gt; &lt;!--[endif]--&gt;

Ask & Answer Question

Physics Class 11th Physics Oscillations

A block ' $A$ ' of mass $m$ attached to a massless spring is performing oscillatory motion of amplitude ' $A$ ' on a frictionless horizontal plane. If it collides elastically with another block ' $B$ ' of mass $2 m$ kept initially at rest, when ' $A$ ' is passing through its equilibrium point, the amplitude of oscillation for the block ' $A$ ' become $f$ A. The value of $f$ is:

Option 1 -

$\frac{1}{\sqrt[]{2}}$

Option 2 -

$\frac{1}{2}$

Option 3 -

Option 4 -

$\frac{1}{3}$

2 Views | Posted 3 weeks ago

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1 Answer

Answered by

Raj Pandey | Contributor-Level 9

3 weeks ago

Correct Option - 4

Detailed Solution:

$v_{m a x} = A ω = A \sqrt[]{\frac{k}{m}}$

Momentum conservation $m v_{max} + 0 = m v_{A} + 2 m v_{B}$

$v_{A} + 2 v_{B} = v_{m a x}$

$v_{2} - v_{1} = u_{1} - u_{2} \Rightarrow v_{B} - v_{A} = v_{m a x}$

By (1) & (2) $v_{A} = - \frac{A ω}{3}$

$|v_{A}| = A_{new} ω$

$\frac{A ω}{3} = A_{new} ω ∴ A_{new} = \frac{A}{3}; f = \frac{1}{3}$

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