question_answer

Two bodies of equal mass are projected horizontally with equal velocity V to move along the paths AB and A' B' as shown in the figure. The path AB and A'B' are frictionless and lie in a vertical plane. Let the time taken by body in making displacement \['d'\] be \[{{t}_{1}}\], and \[{{t}_{2}}\] for path AB and A' B' respectively. (Assume that body does not lose contact at any point). Then

A)  \[{{t}_{1}}={{t}_{2}}\]                   

B)  \[{{t}_{1}}<{{t}_{2}}\]

C)  \[{{t}_{1}}>{{t}_{2}}\]                   

D)  cannot be predicted

Correct Answer: C

Solution :

Path AB                \[a=0\], vel. = constant = V Path A'B'              \[{{a}_{x}}=N\sin \theta \] So, vel. increases up till lowest point. By W.E.T,  \[{{W}_{g}}\] from A' to B' is 0 so \[{{\text{V}}_{\text{f}}}\text{=V}\] From lowest point till B', \[{{a}_{x}}=-N\sin \theta \] Sol vel. decreases up till B' but is always greater than V. Hence, \[{{V}_{A'B'}}>V\]for whole path. \[\therefore \,\,{{t}_{2}}<{{t}_{1}}\]