Question

7.
A bullet of mass 10 g moving horizontally with a
velocity of 400 ms-1 strikes a wood block of mass
2 kg which is suspended by light inextensible
string of length 5 m. As a result, the centre of
gravity of the block found to rise a vertical distance
of 10 cm. The speed of the bullet after it emerges
out horizontally from the block will be
[NEET (Phase-2) 2016]
(1) 100 ms -1
(2) 80 ms-1
(3) 120 ms-1
(4) 160 ms-1
iti
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Answer 1

Answer:

120ms-1 is the answer for this question

Answer 2

Answer:

Mass of the bullet is 10g

Velocity=400m/s

Mass of the wooden block is 2kg

Length=5m

Let the velocity of the bullet and the block after the collision will be v

1

&v

2

The block rises 410cm=0.1m

A/C Conservation of the energy of the block:

PE at the maximum height =KE at the bottom of the block,

m

2

gh=

2

1

m

2

v

2

2

v

2

=

2gh

=

2×9.8×0.1

=1.4m/s

Let u

1

is the initial velocity of the bullet, then according to the conservation of momentum,

m

1

u

1

=m

2

v

2

+m

1

v

1

v

1

=

m

1

m

1

u

1

−m

2

v

2

=

0.01

0.01×400−2×1.4

=120m/s

H