A 20 g bullet travelling with a velocity v becomes embedded into a 4 kg
block of wood suspended from two light strings. If the block-bullet
combination rises to a height of 2 m, what is the initial velocity v of the
bullet ?
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Answers
Explanation:
Let V be velocity of the bullet system just after collision. Then by the law of conservation of linear momentum, we get
mv=(m+M)V
V=
m+M
mv
Let the block rises to a height h.
According to law of conservation of mechanical energy,
we get
2
1
(m+M)V
2
=(m+M)gh
h=
2g
V
2
=
2g
v
2
(
m+M
m
)
Answer:
- Initial velocity of bullet will be approximately 1258.46 m/s
Explanation:
Given,
- Mass of bullet, m = 20 g = (20 / 1000) kg = 0.02 kg
- Initial velocity of bullet it is travelling with = v
- Mass of wooden block, M = 4 kg
- The wooden block is suspended from two light strings. Bullet on firing get embedded into the block. (We will assume strings massless)
- Height to which block-bullet combination is rised, h = 2 m
To find,
- Initial velocity of bullet, v =?
Solution,
Let, the velocity of block-bullet combination just after the collision be V
& the initial velocity of wooden block would be, u = 0
then,
By the law of conservation of linear momentum
→ Initial momentum before collision = Final momentum after collision
→ m v + M u = m V + M V
→ m v + M ( 0 ) = V ( m + M )
→ m v = V ( m + M )
→ V = ( m v ) / ( m + M )
→ V = ( 0.02 v ) / ( 0.02 + 4 )
→ V = ( 0.02 v ) / 4.02 ______equation(1)
Therefore, we have the the velocity of block-bullet combination after collision.
Now,
By the law of conservation of mechanical energy
→ Kinetic energy of system = Potential energy of system
→ 1/2 ( m + M ) V² = ( m + M ) g h
[ where g is acceleration due to gravity ]
→ 1/2 ( 0.02 + 4 ) V² = ( 0.02 + 4 ) ( 9.8 ) ( 2 )
→ V² = 39.2
Using equation (1)
→ [( 0.02 v ) / 4.02]² = 39.2
→ 0.0004 v² / 16.1604 = 39.2
→ v = 1258.46 m/s
Therefore,
- Initial velocity of bullet will be approximately 1258.46 m/s.