A trolley of mass 20 kg carries 16 kg grain and moves on a horizontal smooth and straight track at 20m/s.If the grain starts leaking out of a hole at the bottom at time t=0 s at the rate of .5 kg/s,what will be the speed of trolley at t=22s?
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Answered by
16
after 22 sec mass of trolley &grain=36-(22*.5)
=36-11=25
now from conservation of momentum
(20+16)*20=25*v
speed (v)=(36*20)/25
=36-11=25
now from conservation of momentum
(20+16)*20=25*v
speed (v)=(36*20)/25
Answered by
14
Initial total mass of trolley = 20 kg + 16 kg = 36 kg
Mass of grain leaked in 22 s = 0.5 kg/s × 22 s = 11 s
Total mass of trolley at t = 22 s is
m2 = 20 kg + (16 kg - 11 kg) = 25 kg
Apply conservation of linear momentum
Initial momentum = Final momentum
m1u = m2v
v = m1u / (m2)
= (36 kg × 20 m/s) / (25 kg)
= 28.8 m/s
∴ Speed of trolley at t = 22 s is 28.8 m/s
Mass of grain leaked in 22 s = 0.5 kg/s × 22 s = 11 s
Total mass of trolley at t = 22 s is
m2 = 20 kg + (16 kg - 11 kg) = 25 kg
Apply conservation of linear momentum
Initial momentum = Final momentum
m1u = m2v
v = m1u / (m2)
= (36 kg × 20 m/s) / (25 kg)
= 28.8 m/s
∴ Speed of trolley at t = 22 s is 28.8 m/s
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