Question bank/JEE Main/Physics

A system is shown in the figure. Block A is moving with 1 m /s
towards left. Wedge is moving with 1 m /s towards right. Then speed of
the block B will be

A1 m /sB2 m /sCnone of these

Show answer and solution▾

Velocity of block A w.r.t. wedge is 2 m So we have
${\overset{\rightarrow}{V}}_{BW} = {\overset{\rightarrow}{V}}_{B} - {\overset{\rightarrow}{V}}_{W} \Rightarrow {\overset{\rightarrow}{V}}_{B} = {\overset{\rightarrow}{V}}_{BW} + {\overset{\rightarrow}{V}}_{W}$
So
$V_{B} = \sqrt{{(V_{W})}^{2} + {(V_{BW})}^{2} + 2V_{W} \times V_{BW}(cos120^{\circ})}$
$= \sqrt{1^{2} + 2^{2} + 2 \times 1 \times 2 \times ( - 1/2)} = \sqrt{3}m/s$

Source: Grass-G / P-18 Physics paper 27 Nov Third.docx