A bullet of mass 0.01 kg and travelling at a speed of 500 m/s strikes a block of 2 kg which is suspended by a string of length 5m. The centre of gravity of the block is found to rise a vertical distance of 0.1m. What is the speed of the bullet after it emerges from the block?
a)
200 m/s
b)
220 m/s
c)
204 m/s
d)
284 m/s
A bullet of mass 0.01 kg and travelling at a speed of 500 m/s strikes a block of 2 kg which is suspended by a string of length 5m. The centre of gravity of the block is found to rise a vertical distance of 0.1m. What is the speed of the bullet after it emerges from the block?
a)
200 m/s
b)
220 m/s
c)
204 m/s
d)
284 m/s
Question ID - 100190 | SaraNextGen Answer
A bullet of mass 0.01 kg and travelling at a speed of 500 m/s strikes a block of 2 kg which is suspended by a string of length 5m. The centre of gravity of the block is found to rise a vertical distance of 0.1m. What is the speed of the bullet after it emerges from the block?
a)
200 m/s
b)
220 m/s
c)
204 m/s
d)
284 m/s
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A bullet of mass 0.01 kg and travelling at a speed of 500 m/s strikes a block of 2 kg which is suspended by a string of length 5m. The centre of gravity of the block is found to rise a vertical distance of 0.1m. What is the speed of the bullet after it emerges from the block? |
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a) |
200 m/s |
b) |
220 m/s |
c) |
204 m/s |
d) |
284 m/s |
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(b) Suppose the velocity of the bullet of mass Applying law of conservation of linear momentum, we get
Or 500 Or 5 = 2 Now, the kinetic energy gained by the block
Or Putting the value of
|
is 
and it strikes the block of mass
. After collision, the linear velocity of the block is 
and that of the bullet is
0.01 = 2
+0.01
+ 0.01
(i)
raises it to a height 
where it gains gravitational potential energy 
By conservation of energy, we get
m/s
in Eq. (i), we get
or
m/s