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A particle moves in the X-Y plane under the influence of a force such that it's linear momentum is P(t)= A[ i cos(kt) - j sin(kt)] , where A and k are constants. The angle between the force and the momentum is (a) 0 (b) 30 (c) 45 (d) 90?
Verified Answer
A particle moves in the X-Y plane under the influence of a force such ...
Ans.
Option (d)
P(t) = A [j˄
cos
kt
– j˄ sin
kt
]
F = [{
dp
(t)} / {
dt
}]
F = (d /
dt
) A [
i
˄
cos
kt
– j˄ sin
kt
]
= A [–
i
˄ sin
kt
(k)– j
cos
kt
(k)]
F =
Ak
[–
i
˄ sin
kt
– j˄
cos
kt]
angle between two vectors is θ
cos
θ = {(
F ∙
P) / (|
F| |
P|)}
= A [{(j˄
cos
kt
– j˄ sin
kt
) ∙
Ak
(–
i
˄ sin
kt
– j˄
cos
kt
)} / {|
F||
P|}]
= A
2 k [{– sin
kt
cos
kt
+ sin
kt
cos
kt
} / {|
F| |
P|}]
= A
2k(0)
= 0
θ = cos
–1(0)
θ = 90�
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Most Upvoted Answer
A particle moves in the X-Y plane under the influence of a force such ...
Analysis:
To determine the angle between the force and the momentum, we need to find the dot product of the force and momentum vectors. The dot product is given by the formula:
P(t) · F(t) = |P(t)||F(t)|cosθ
where P(t) is the momentum vector, F(t) is the force vector, and θ is the angle between them.
Finding the Force Vector:
The force vector can be determined by taking the derivative of the momentum vector with respect to time. Differentiating P(t) with respect to t, we get:
dP(t)/dt = -A[ i k sin(kt) + j k cos(kt)]
This is the force vector F(t).
Finding the Dot Product:
Now, we can find the dot product of the momentum and force vectors:
P(t) · F(t) = A[ i cos(kt) - j sin(kt)] · [-A( i k sin(kt) + j k cos(kt))]
Simplifying the dot product, we get:
P(t) · F(t) = -A^2[k cos^2(kt) + k sin^2(kt)]
P(t) · F(t) = -A^2k[cos^2(kt) + sin^2(kt)]
P(t) · F(t) = -A^2k
Angle between Force and Momentum:
The dot product of two vectors is equal to the product of their magnitudes and the cosine of the angle between them. Therefore, from the dot product equation:
P(t) · F(t) = |P(t)||F(t)|cosθ
we can see that if the dot product is negative, the angle between the vectors is 90 degrees (option d). If the dot product is positive, the angle is 0 degrees (option a). If the dot product is zero, the angle is 90 degrees (option d).
From the dot product we found earlier:
P(t) · F(t) = -A^2k
Since A and k are positive constants, the dot product is negative. Therefore, the angle between the force and momentum is 90 degrees (option d).
Conclusion:
The angle between the force and momentum is 90 degrees (option d).
To determine the angle between the force and the momentum, we need to find the dot product of the force and momentum vectors. The dot product is given by the formula:
P(t) · F(t) = |P(t)||F(t)|cosθ
where P(t) is the momentum vector, F(t) is the force vector, and θ is the angle between them.
Finding the Force Vector:
The force vector can be determined by taking the derivative of the momentum vector with respect to time. Differentiating P(t) with respect to t, we get:
dP(t)/dt = -A[ i k sin(kt) + j k cos(kt)]
This is the force vector F(t).
Finding the Dot Product:
Now, we can find the dot product of the momentum and force vectors:
P(t) · F(t) = A[ i cos(kt) - j sin(kt)] · [-A( i k sin(kt) + j k cos(kt))]
Simplifying the dot product, we get:
P(t) · F(t) = -A^2[k cos^2(kt) + k sin^2(kt)]
P(t) · F(t) = -A^2k[cos^2(kt) + sin^2(kt)]
P(t) · F(t) = -A^2k
Angle between Force and Momentum:
The dot product of two vectors is equal to the product of their magnitudes and the cosine of the angle between them. Therefore, from the dot product equation:
P(t) · F(t) = |P(t)||F(t)|cosθ
we can see that if the dot product is negative, the angle between the vectors is 90 degrees (option d). If the dot product is positive, the angle is 0 degrees (option a). If the dot product is zero, the angle is 90 degrees (option d).
From the dot product we found earlier:
P(t) · F(t) = -A^2k
Since A and k are positive constants, the dot product is negative. Therefore, the angle between the force and momentum is 90 degrees (option d).
Conclusion:
The angle between the force and momentum is 90 degrees (option d).
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A particle moves in the X-Y plane under the influence of a force such that it's linear momentum is P(t)= A[ i cos(kt) - j sin(kt)] , where A and k are constants. The angle between the force and the momentum is (a) 0 (b) 30 (c) 45 (d) 90?
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A particle moves in the X-Y plane under the influence of a force such that it's linear momentum is P(t)= A[ i cos(kt) - j sin(kt)] , where A and k are constants. The angle between the force and the momentum is (a) 0 (b) 30 (c) 45 (d) 90? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A particle moves in the X-Y plane under the influence of a force such that it's linear momentum is P(t)= A[ i cos(kt) - j sin(kt)] , where A and k are constants. The angle between the force and the momentum is (a) 0 (b) 30 (c) 45 (d) 90? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle moves in the X-Y plane under the influence of a force such that it's linear momentum is P(t)= A[ i cos(kt) - j sin(kt)] , where A and k are constants. The angle between the force and the momentum is (a) 0 (b) 30 (c) 45 (d) 90?.
A particle moves in the X-Y plane under the influence of a force such that it's linear momentum is P(t)= A[ i cos(kt) - j sin(kt)] , where A and k are constants. The angle between the force and the momentum is (a) 0 (b) 30 (c) 45 (d) 90? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A particle moves in the X-Y plane under the influence of a force such that it's linear momentum is P(t)= A[ i cos(kt) - j sin(kt)] , where A and k are constants. The angle between the force and the momentum is (a) 0 (b) 30 (c) 45 (d) 90? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle moves in the X-Y plane under the influence of a force such that it's linear momentum is P(t)= A[ i cos(kt) - j sin(kt)] , where A and k are constants. The angle between the force and the momentum is (a) 0 (b) 30 (c) 45 (d) 90?.
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