Question

A point like object of mass M is thrown up from point A as shown in the figure. It slides along the full length of the smooth track ABC (of radius R for part BC). Let R=1 m,AB=2 m,M=0.5 kg,g=10 m/s2 & OD=3m.

A
Minimum speed V0 to slide full length of the track is 60m/s
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B
Minimum speed V0 to reach D is 105m/s
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C
Normal force on the object by the track at C if it reaches D is 15 N
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D
Normal force on the object by the track at B if it reaches D is 32.5 N
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Solution

The correct option is D Normal force on the object by the track at B if it reaches D is 32.5 NFor the object to slide full length, its normal reaction at C should be greater than zero. Nc=mV2cR−mg Nc>0⇒Vc>√Rg By conservation of energy, ⇒12mV20=mg×3+12×m×V2c ⇒12mV20=mg×3+12×m Rg V20=6g+Rg=7g⇒Vo=√70m/s Consider the case where the body falls exactly at D. By Sy=uyt+12ayt2 1=0+12gt2⇒t=1√5 And, Sx=uxt+12axt2⇒3=Vc1√5 So Vc=3√5 By conservation of energy, ⇒12m/V20=m/g×3+12×m/.(3√5)2 V20=60+45⇒V0=√105m/s Similarly by conservation of energy, ⇒VB=√65 NB=mV2BR=0.5×651=32.5N

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