P5. A bullet with an initial velocity of 280 m/s in the x-direction penetrates an initially stationary block of mass 11 kg and e
1 answer:
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We can solve the problem by applying Newton's second law, which states that the resultant of the forces acting on an object is equal to the product between its mass and its acceleration:
We should consider two different directions: the direction perpendicular to the inclined plane and the direction parallel to it. Let's write the equations of the forces along the two directions, decomposing the weight of the object (mg):
(parallel direction) (1)
(perpendicular direction) (2)
where
is the angle of the inclined plane, N is the normal reaction of the plane, 
is the frictional force, with 
being the coefficient of friction.
From eq.(2), we find
and if we substitute into eq.(1), we can find the acceleration of the block:
from which
We should consider two different directions: the direction perpendicular to the inclined plane and the direction parallel to it. Let's write the equations of the forces along the two directions, decomposing the weight of the object (mg):
where
From eq.(2), we find
and if we substitute into eq.(1), we can find the acceleration of the block:
from which
Answer:
Maximum height attained by the model rocket is 2172.87 m
Explanation:
Given,
- Initial speed of the model rocket = u = 0
- acceleration of the model rocket =
- time during the acceleration = t = 2.30 s
We have to consider the whole motion into two parts
In first part the rocket is moving with an acceleration of a = 85.0 for the time t = 2.30 s before the fuel abruptly runs out.
Let be the height attained by the rocket during this time intervel,
And Final velocity at that point be v
Now, in second part, after reaching the altitude of 224.825 m the fuel abruptly runs out. Therefore rocket is moving upward under the effect of gravitational acceleration,
Let '' be the altitude attained by the rocket to reach at the maximum point after the rocket's fuel runs out,
At that insitant,
- initial velocity of the rocket = v = 195.5 m/s.
- a =
- Final velocity of the rocket at the maximum altitude =
From the kinematics,
Hence the maximum altitude attained by the rocket from the ground is