NEET-XI-Physics
exam-2 year:2016
- #117A bullet of mass 10g moving horizontally with a
velocity of 400 ``ms^{-1}`` strikes a wooden block of mass
2 kg which is suspended by a light inextensible string
of length 5 m. As a result, the centre of gravity of
the block is found to rise a vertical distance of
10 cm. The speed of the bullet after it emerges out
horizontally from the block will be :-
(1) 120 ``ms^{-1}``
(2) 160 ``ms^{-1}``
(3) 100 ``ms^{-1}``
(4) 80 ``ms^{-1}``digAnsr: 1Ans : (1)
Sol.
2kg
2kg v1
v210gm
400 m/s
0.1 m
5m
Applying momentum conservation
1 2
10 10
400 0 2 v v
1000 1000
ï‚´ + = ï‚´ + ï‚´
&implies; 4 = 2v1 + 0.01v2 ......(1)
Applying work energy theorem for block
W = â–µKE
&implies; 2 × 10 × 0.1 =
1
2
× 2 × v12
&implies; v1 = 2 = 1.4 m/s
Putting the value of v1 in equation (1)
4 = 2 × 1.4 + 0.01 v2 &implies;ï€ v2 = 120 m/s