A body of mass 10 kg and velocity 10 m/s collides with a stationary body of mass 5 kg . After collision both bodies stick to each other , velocity of the bodies after collision will be

Options

A. a) 3 /10 m/s
B. b) 18 /3 m/s
C. c) 9 /2 0 m/s
D. d) 20 /3 m/s

Explanation

(a) In the circuit shown if we assume that potential at A is zero then potential at A is zero then potential at B is `epsi-ir`. Now since the connecting wires are of zero resistance
`thereforeV_(D)=V_(A)=0rArrV_(C)=B_(B)=epsi-ri`
Now current through CD is also i (because it's in series with the cell).
`therefore i=(V_(C)-V_(D))/R=((epsi-ir)-0)/R" Current i"=epsi/(r+R)`
Note : After learning the concept of series combination we will be able to calculate the current directly
(b) Power output `P=i^(2)R=epsi^(2)/((r+R)^(2))R`
( c) `(dp)/(dR)=epsi^(2)/((r+R)^(2))-(2epsi^(2)R)/((r+R)^(3))[R+r-2R]`

for maximum power supply `(dp)/(dR)=0`
`rArr r+R-2R=0rArrr=R`
Here for maximum power output outer resistance should be equal to internal resistance
(d) `P_("max")=epsi^(2)/(4r)`
( e) Graph between 'P' and R maximum power output at R=r
`P_("max")=epsi^(2)/(4r)rArri=epsi/(r+R)`
(f) Power output `P=(epsi^(3)R)/((r+R)^(2))`
`P(r^(2)+2rR+R^(2))=epsi^(2)R`
`R^(2)+(2r-epsi^(2)/P)R+r^(2)=0`
Above quadratic equation in R has two roots `R_(1)" and "R_(2)` for given values of `epsi,` P and r such that
`thereforeR_(1)R_(2)=r_(2)("product of roots")`
`r^(2)=R_(1)R_(2)`
(g) Power of battery spent `=epsi^(2)/((r+)^(2)),2r=epsi^(3)/(2r)`
Power (output)`=(epsi/(r+r))^(2)xxr=epsi^(2)/(4r)`
`"Efficiency ="("Power output")/("total power by cell")=(epsi^(2)/(4r)xx100)/(epsi^(2)/(2r))=1/2xx100=50%`