1.

F particle moves in x-y plane under the action of force vecF such that the value of its linear momentum (vecp) at any instant t is p_x=2 cos t and py=2 sin t. The angle theta between vecFvecP at a given time t will be-

Answer»

`90^(@)`
`0^(@)`
`180^(@)`
`30^@`

Solution :`p=sqrt(px^(2)+PY^(2))=sqrt(4 COS^(2)t+4 sin^(2)t)=2` (constant), As p remains constant, `vecF` ACTS at right angle to `vecp`.


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