The vector obtained by addition of two vectors is termed as ______
a) New vector
b) Resultant vector
c) Derived vector
d) Sum vector

Answer

A, B, C, D

Explanation

a. Net force on the particle A and B, `F=mgsin theta`
work done to reach B,
`W=FS=mg sintheta l`
Again work done to reach B to C
`=mgh`
`=mghR(1-costheta)`
`So, total work done
`=mg(lsintheta+r(1-costheta)]`

Now, change in K.E.
=work done
`rarr 1/2mv_2^2=mg[lsintheta+R(1-costheta)]`
`rarr v_2=sqrt(2g[R(1-costheta)+lsintheta]`
b. Whenteh block is projected at a speed `2v_0`

Let the velocity of C will be `v_0`.
Applying energyy priciple
`(1/2)mv_0^2 -(1/2)m(2v_0)^2`
=-mg[lsintheta+R(1-costheta)]`
rarr V^2=4v_0^2-2g[lsintheta+R(1-costheta)]`
`=4.2g[lsintheta+R(1-costheta)]- 2g[lsintheta+R(1-costeta)]`
so force acting on teh body
`rarr N=V^2/R=6mg[(l/R)sintheta+1-costheta]`
c. Let the loose contact after making an angle `theta`
`(mv^2)/R=mgcostheta`
`rarr v^2=Rgcostheta` .......i
`Again `1/2 mv^2=mg(R-Rcostheta)`
`rarr v^2=2gR(1-costheta)`.........ii
`Form i and ii `costheta =2/3
`rarr theta=cos^-1(2/3)`