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Question

# A car at rest on a horizontal surface (with coefficient of friction 0.1) has total mass 50 kg. Gases are ejected from this car backwards with relative velocity 20 m/s. The rate of ejection of gases is 2 kg/s. Total mass of gas is 20 kg. Find the maximum speed of the car in (m/s) Take g=10 m/s2,ln(43)=0.29

A
0.2 m/s
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B
0.8 m/s
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C
0.6 m/s
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D
0.4 m/s
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Solution

## The correct option is B 0.8 m/sGiven, Total mass of the car (m0)=50 kg Total mass of the gas (m)=20 kg Relative velocity of ejection of gases vr=20 m/s & Rate of ejection =2 kg/s Car will start moving when the thrust force overcomes the limiting force of friction. i.e μmg=vr(−dmdt) 0.1×(50−2t)×10=20×2 (∵m=m0−μt) ⇒t=5 s Therefore, car will start moving after t=5 seconds At t=5 s, mass of the car will be 50−2×5=40 kg Total time taken to consume fuel (t)=10 s At t=10 s, mass of the car will be 30 kg Assuming t=0 when the car starts moving, this means max velocity will be achieved at t=10−5=5 s Thus, final velocity (max) of the car will be v=vrln(m0m)−μgt vmax=20×ln(4030)−0.1×10×(10−5)=0.80 m/s

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