The correct option is
C Velocity of trolley is
1003 m/sGiven, initial mass of trolley,
m0=2 kg
Rate of mass addition for trolley,
dmdt=μ=0.1 kg/s
Final mass of trolley after
10 s is
m=m0+μt, where
t=10 s
⇒m=2+0.1×10
⇒m=3 kg
Let the velocity of the trolley be
v at
t=10 s. Velocity of sand is zero as it is poured gently into the trolley.
⇒ Velocity of sand w.r.t trolley is,
vst=vs−vt=0−v=−v
The velocity of sand w.r.t trolley is in left direction and mass is added, therefore thrust force
(Ft) will act in the direction of
vst i.e towards left.
Hence, net force on trolley,
Fnet=F−Ft
⇒Fnet=F−vdmdt
⇒ma=F−vdmdt
Substituting the values of
m & v in the above equation,
⇒(m0+μt)dvdt=F−μv ...(1)
⇒v∫0dvF−μv=t∫0dtm0+μt
Applying limits of velocity
v=0→v and time
t=0→t to get the velocity of trolley at any instant of time
t
⇒[ln(F−μv)−μ]v0=[ln(m0+μt)μ]t0
⇒[−ln(F−μv)+lnF]=[ln(m0+μt)−lnm0]
Applying property,
lna−lnb=lnab
⇒lnFF−μv=lnm0+μtm0
⇒FF−μv=m0+μtm0
Substituting values of
m0=2 kg, μ=0.1 kg/s, t=10 s, F=10 N
⇒1010−0.1v=2+0.1×102
⇒v=1003 m/s will be the velocity of trolley at
t=10 s.
Hence, net force on trolley will be,
Fnet=F−Ft
⇒Fnet=10−(0.1×1003)=203 N will be the net force on the trolley at
t=10 s.
∴ options (a), (b), (c) are correct.