Mass A is released from rest the of a frictionless inclined plane 18 m long and reaches the bottom 3 s later- At the instant when A is released- a second mass B is projected upwards along the plate from the bottom with a certain initial velocity- Mass B travels a distance up the plane- stops and returns to the bottom so that it arrives simultaneously with A- The two masses do not collide- Initial velocity of B is4-ms-1-6-ms-1-7-ms-1-5-ms-1-

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Toppr Admin · Accepted Answer

Here- for A-18-0-xD7-3-12a-xD7-32 or a-4ms-x2212-2 For B- time taken to move up is given by - t1-u-a -x2234- the relation v-u-at here becomes 0-u-x2212-at1-xA0-Distance moved up is given by the relation 0-u2-x2212-2aSi-e S-u2-2a

Here, for A,18=0×3+12a×32 or a=4ms−2 For B, time taken to move up is given by , t1=u/a (∴ the relation v=u+at here becomes 0=u−at1). Distance moved up is given by the relation 0=u2−2aSi.e S=u2/2a

Here, for $A, 18 = 0 \times 3 + \frac{1}{2} a \times 3^{2}$ or $a = 4 m s^{- 2}$
For $B$ , time taken to move up is given by , $t_{1} = u / a$ ( $∴$ the relation $v = u + a t$ here becomes $0 = u - a t_{1}$ ).
Distance moved up is given by the relation $0 = u^{2} - 2 a S$
i.e $S = u^{2} / 2 a$