A bullet of mass 0.01kg and travelling at a speed of 500 m/s strikes a block of mass 2kg which is suspended by a string of length 5m. The centre of gravity of the block is found to raise a vertical distance of 0.2m. What is the speed of the bullet after it emerges from the block?
Answers
Explanation:
PHYSICS
A bullet of mass 0.01kg and traveling at a speed of 500m/s strikes and passes horizontally through a block of mass 2kg which is suspended by a string of length 5m. The center of gravity of the block is found to raise a vertical distance of 0.1m. What is the speed of the bullet after it emerges from the block (g=9.8m/s
2
) (time of the passing of bullet is negligible)
December 26, 2019avatar
Brijraj Balla
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ANSWER
Given: A bullet of mass 0.01kg and traveling at a speed of 500m/s strikes and passes horizontally through a block of mass 2kg which is suspended by a string of length 5m. The center of gravity of the block is found to raise a vertical distance of 0.1m.
To find the speed of the bullet after it emerges from the block
Solution:
Suppose v
1
and v
2
are the velocities of the bullet and the block after collision.
Since the block rises to a height of h=0.1m, so all its kinetic energy is converted into its potential enegy
Thus by conservation of energy
2
1
m
2
v
2
2
=m
2
gh
⟹v
2
=
2gh
Substituting the corresponding values we get
v
2
=
2×9.8×0.1
⟹v
2
=1.4m/s
If u
1
is the initial velocity of the bullet, then applying the law of conservation of momentum along the initial direction of the bullet, we get
m
1
u
1
=m
1
v
1
+m
2
v
2
⟹v
1
=
m
1
m
1
u
1
−m
2
v
2
Now substituting the values given , we get
⟹v
1
=
0.01
0.01×500−2×1.4
⟹v
1
=220m/s
As there is no external force between bullet and block so, momentum will remain conserved.....
Initial momentum,of bullet=.01×500=5
initial momentum of block=2×0=0
total, initial momentum=5
when the bulllet collides with block
then,
intial velocity in bullet=v1(say)
acceleration=g
final velocity=0
distance travelled=.2
using Newton's third law of motion
0^2=v1^2-2×10×.2
V1^2=4
V1=2
final momentum in block=V1×M=2×2=4
final momentum in bullet=.01×v
total final momentum=4+.01v
total final momentum=total initial momentum
5=4+.01v
v=100m/s
Hence the velocity of the bulllet will be 100m/s,when it will emerge out from the block