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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

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

Grade:11

2 Answers

Agrata Singh
208 Points
5 years ago
 
 
Writing energy conservation for the system just before collision and when the bag reaches height of 10 cm
Einitial= 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=402
v=20(approx) 
 
Suraj
13 Points
5 years ago
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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