Two masses A and B, each of mass M are fixed together by a massless spring. A force acts on the mass B as shown in figure. If the mass A starts moving away from mass B with acceleration 'a', then the acceleration of mass B will be:

Option 1 - <math> <mrow> <mfrac> <mrow> <mi>M</mi> <mi>a</mi> <mo>−</mo> <mi>F</mi> </mrow> <mrow> <mi>M</mi> </mrow> </mfrac> </mrow> </math>

Option 2 - <math> <mrow> <mfrac> <mrow> <mi>M</mi> <mi>F</mi> </mrow> <mrow> <mi>F</mi> <mo>+</mo> <mi>M</mi> <mi>a</mi> </mrow> </mfrac> </mrow> </math>

Option 3 - <math> <mrow> <mfrac> <mrow> <mi>F</mi> <mo>+</mo> <mi>M</mi> <mi>a</mi> </mrow> <mrow> <mi>M</mi> </mrow> </mfrac> </mrow> </math>

Option 4 - <math> <mrow> <mfrac> <mrow> <mi>F</mi> <mo>−</mo> <mi>M</mi> <mi>a</mi> </mrow> <mrow> <mi>M</mi> </mrow> </mfrac> </mrow> </math>

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Alok Kumar Singh | Contributor-Level 10

4 months ago

Correct Option - 4

Detailed Solution:

Let the spring is in extension state and $a_{B} > a_{A}, {\vec{a}}_{A B} = - a_{A} \hat{i} - (- a_{B} \hat{i}) = (- a_{A} + a_{B}) \hat{i}$

Hence we can say that block moves away from block B in the frame of B

F – kx = Ma_B . (1)

kx = Ma . (2)

$\Rightarrow a_{B} = \frac{F}{M} - a$

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