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Question

Sand is poured gently at a constant rate of 0.1 kg/s on the trolley of initial mass 2 kg kept over a smooth surface at rest. The external force acting on the trolley is 10 N as shown in the figure. Then, identify the correct statement(s) for the trolley after time 10 sec:


A
Mass of trolley is 3 kg
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B
Net force on trolley is 203 N
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C
Velocity of trolley is 1003 m/s
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D
Velocity of trolley is 2003 m/s
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Solution

The correct option is C Velocity of trolley is 1003 m/s
Given, 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=vsvt=0v=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=FFt
Fnet=Fvdmdt
ma=Fvdmdt
Substituting the values of m & v in the above equation,
(m0+μt)dvdt=Fμv ...(1)
v0dvFμv=t0dtm0+μt
Applying limits of velocity v=0v and time t=0t to get the velocity of trolley at any instant of time t
[ln(Fμv)μ]v0=[ln(m0+μt)μ]t0
[ln(Fμv)+lnF]=[ln(m0+μt)lnm0]

Applying property, lnalnb=lnab
lnFFμv=lnm0+μtm0
FFμv=m0+μtm0
Substituting values of m0=2 kg, μ=0.1 kg/s, t=10 s, F=10 N
10100.1v=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=FFt
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.

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