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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Physics
Chapter
Laws of Motion

BITSAT 2015 · Physics · Laws of Motion

The masses of blocks A and B are m and M respectively. Between A...

The masses of blocks A\mathrm{A} and B\mathrm{B} are m\mathrm{m} and M\mathrm{M} respectively. Between A\mathrm{A} and B\mathrm{B}, there is a constant frictional force F\mathrm{F} and B\mathrm{B} can slide on a smooth horizontal surface. A is set in motion with velocity while B\mathrm{B} is at rest. What is the distance moved by A relative to B before they move with the same velocity? The masses of blocks A and B are m and M respectively. Between A and B , there is a constant frictional force F and B

Options

  1. AmMv02F(m−M)\frac{\mathrm{mMv}_{0}^{2}}{\mathrm{ F}(\mathrm{ m}-\mathrm{M})}
  2. BmMv022F(m−M)\frac{\mathrm{mMv}_{0}^{2}}{2 \mathrm{ F}(\mathrm{ m}-\mathrm{M})}
  3. CmMv02F(m+M)\frac{\mathrm{mMv}_{0}^{2}}{\mathrm{ F}(\mathrm{ m}+\mathrm{M})}
  4. DmMv022F(M+m)\frac{\mathrm{mMv}_{0}^{2}}{2 \mathrm{ F}(\mathrm{M}+\mathrm{m})}

Correct answer

D. mMv022F(M+m)\frac{\mathrm{mMv}_{0}^{2}}{2 \mathrm{ F}(\mathrm{M}+\mathrm{m})}

Step-by-step solution

For the blocks A and B FBD as shown below The masses of blocks A and B are m and M respectively. Between A and B , there is a constant frictional force F and B
Equations of motion aA=FM( in −x direction )aB=FM( in +x direction) \begin{array}{l}\mathrm{a}_{\mathrm{A}}=\frac{\mathrm{F}}{\mathrm{M}}(\text{ in }-\mathrm{x} \text{ direction })\\ \mathrm{a}_{\mathrm{B}}=\frac{\mathrm{F}}{\mathrm{M}}(\text{ in }+\mathrm{x} \text{ direction) } \end{array} Relative acceleration, of A w.r.t. B, aA,B=aA−aB=−Fm−FM=−F(M+mMm)( along −x direction )\begin{aligned}\mathrm{a}_{\mathrm{A}, \mathrm{B}}&=\mathrm{a}_{\mathrm{A}}-\mathrm{a}_{\mathrm{B}}=-\frac{\mathrm{F}}{\mathrm{m}}-\frac{\mathrm{F}}{\mathrm{M}}\\ &=-\mathrm{F}\left(\frac{\mathrm{M}+\mathrm{m}}{\mathrm{Mm}}\right)(\text{ along }-\mathrm{x} \text{ direction }) \end{aligned} Initial relative velocity of A w.r.t. B, uAB=v0\mathrm{u}_{\mathrm{AB}}=\mathrm{v}_{0} using equation v2=u2+2v^{2}=u^{2}+2 as 0=v_{0}^{2}-2F(m+M)SMm⇒S=Mmv022F(m+M)\frac{2 \mathrm{ F}(\mathrm{ m}+\mathrm{M}) \mathrm{S}}{\mathrm{Mm}} \Rightarrow \mathrm{S}=\frac{\mathrm{Mmv}_{0}^{2}}{2 \mathrm{ F}(\mathrm{ m}+\mathrm{M})} i.e., Distance moved by A relative to B SAB=Mmv022F(m+M)\mathrm{S}_{\mathrm{AB}}=\frac{\mathrm{Mmv}_{0}^{2}}{2 \mathrm{ F}(\mathrm{ m}+\mathrm{M})}
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