A piece of wood of mass 0.03 kg is dropped from the top of a mass 0.02 kg is fired vertically upward, with a velocity 100
, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is : (g=10
)
(a)
30m
(b)
10m
(c)
40m
(d)
20m
A piece of wood of mass 0.03 kg is dropped from the top of a mass 0.02 kg is fired vertically upward, with a velocity 100 , from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is : (g=10 )
(a)
30m
(b)
10m
(c)
40m
(d)
20m
Question ID - 50231 | SaraNextGen Top Answer
A piece of wood of mass 0.03 kg is dropped from the top of a mass 0.02 kg is fired vertically upward, with a velocity 100 , from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is : (g=10 )
(a)
30m
(b)
10m
(c)
40m
(d)
20m
|
A piece of wood of mass 0.03 kg is dropped from the top of a mass 0.02 kg is fired vertically upward, with a velocity 100 |
|||
|
(a) |
30m |
(b) |
10m |
|
(c) |
40m |
(d) |
20m |
Time taken for the particles to collide,
T=
=
=1sec
Speed of wood just before collision=gt=10m/s & and speed of bullet just before collision v−gt =100−10=90 m/s
Now, conservation of linear momentum just before and after the collision
(0.02) (1v) + (0.02) (9v)= (0.05)v
⇒150 = 5v
⇒ v = 30m/s
Max. height reached by body h=
h=
=45m
Height above tower =40m