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A bag of sand of mass 2 kg is suspended by a rope. A bullet of mass 10g is fired at it and gets embedded into it. The bag rises up a vertical height of 10 cm. The initial velocity of the bullet is nearly- a) 70 m/s b) 140 m/s c) 210 m/s d) 280 m/s
Writing energy conservation for the system just before collision and when the bag reaches height of 10 cmEinitial= 1/2x0.01v2( only the bullet had kinetic energy)Efinal=(2+0.01)x0.1x10 (potential energy of bag+bullet at height 10 cm)Upon solving Einitial=Efinal you will get V2=402v=20(approx)
By conservation of energy,1/2mv^2=mgh (m gets cancelled) 1/2v^2= 10 × 0.01 (Taking g=10m/s^2, h=0.01m V=√0.2-------------1 By conservation of momentum, Mass of bullet × velocity of bullet = (Mass of bullet +mass of bag) × velocity of bag 0.01kg × velocity of bullet = (0.01 + 2) × √0.2 Velocity of bullet= 2.75/0.01 = 280(approx)
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