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Two particle ab are point - 2 m - 2 m and 2 m - 2 m at t = 0 and moving with velocity - 2 I cap 2 JK per second to ICAP 2 JK m per second respectively the closes distance of approach and time when this happen respectively are?
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Two particle ab are point - 2 m - 2 m and 2 m - 2 m at t = 0 and movin...
Given information:
- Particle A is located at (-2m, -2m) and is moving with a velocity of (-2i + 2j) m/s.
- Particle B is located at (2m, -2m) and is moving with a velocity of (i + 2j) m/s.
Approach:
We can use the concept of relative motion to find the closest distance of approach and the time when it happens.
Step 1: Find the relative velocity
The relative velocity of particle B with respect to particle A is given by the vector subtraction of their velocities:
Vrel = Vb - Va
= (i + 2j) - (-2i + 2j)
= 3i
Step 2: Find the equations of motion
We can consider the motion of particle A as the reference frame. The equation of motion for particle B with respect to A is given by:
r = r0 + Vrel * t
Step 3: Solve for the closest distance of approach
To find the closest distance of approach, we need to minimize the magnitude of the relative position vector r. This occurs when the relative position vector is perpendicular to the relative velocity vector.
Let the position vector of particle B with respect to A be r = x * i + y * j.
The dot product of r and Vrel should be zero:
r · Vrel = x * 3 = 0
=> x = 0
Therefore, the closest distance of approach is when particle B is directly above particle A.
Step 4: Find the time of closest approach
Substituting x = 0 in the equation of motion, we have:
r = r0 + Vrel * t
0i + yj = (2i - 2j) + 3it
Comparing the coefficients of i and j, we get:
2 = 3t
-2 = y
Solving these equations, we find:
t = 2/3 s
y = -2 m
Therefore, the time of closest approach is t = 2/3 s, and the closest distance of approach is 2 m.
Summary:
The closest distance of approach between particle A and B is 2 m, and it occurs at t = 2/3 s. At this time, particle B is directly above particle A.
- Particle A is located at (-2m, -2m) and is moving with a velocity of (-2i + 2j) m/s.
- Particle B is located at (2m, -2m) and is moving with a velocity of (i + 2j) m/s.
Approach:
We can use the concept of relative motion to find the closest distance of approach and the time when it happens.
Step 1: Find the relative velocity
The relative velocity of particle B with respect to particle A is given by the vector subtraction of their velocities:
Vrel = Vb - Va
= (i + 2j) - (-2i + 2j)
= 3i
Step 2: Find the equations of motion
We can consider the motion of particle A as the reference frame. The equation of motion for particle B with respect to A is given by:
r = r0 + Vrel * t
Step 3: Solve for the closest distance of approach
To find the closest distance of approach, we need to minimize the magnitude of the relative position vector r. This occurs when the relative position vector is perpendicular to the relative velocity vector.
Let the position vector of particle B with respect to A be r = x * i + y * j.
The dot product of r and Vrel should be zero:
r · Vrel = x * 3 = 0
=> x = 0
Therefore, the closest distance of approach is when particle B is directly above particle A.
Step 4: Find the time of closest approach
Substituting x = 0 in the equation of motion, we have:
r = r0 + Vrel * t
0i + yj = (2i - 2j) + 3it
Comparing the coefficients of i and j, we get:
2 = 3t
-2 = y
Solving these equations, we find:
t = 2/3 s
y = -2 m
Therefore, the time of closest approach is t = 2/3 s, and the closest distance of approach is 2 m.
Summary:
The closest distance of approach between particle A and B is 2 m, and it occurs at t = 2/3 s. At this time, particle B is directly above particle A.
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Two particle ab are point - 2 m - 2 m and 2 m - 2 m at t = 0 and moving with velocity - 2 I cap 2 JK per second to ICAP 2 JK m per second respectively the closes distance of approach and time when this happen respectively are?
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Two particle ab are point - 2 m - 2 m and 2 m - 2 m at t = 0 and moving with velocity - 2 I cap 2 JK per second to ICAP 2 JK m per second respectively the closes distance of approach and time when this happen respectively are? 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 Two particle ab are point - 2 m - 2 m and 2 m - 2 m at t = 0 and moving with velocity - 2 I cap 2 JK per second to ICAP 2 JK m per second respectively the closes distance of approach and time when this happen respectively are? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two particle ab are point - 2 m - 2 m and 2 m - 2 m at t = 0 and moving with velocity - 2 I cap 2 JK per second to ICAP 2 JK m per second respectively the closes distance of approach and time when this happen respectively are?.
Two particle ab are point - 2 m - 2 m and 2 m - 2 m at t = 0 and moving with velocity - 2 I cap 2 JK per second to ICAP 2 JK m per second respectively the closes distance of approach and time when this happen respectively are? 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 Two particle ab are point - 2 m - 2 m and 2 m - 2 m at t = 0 and moving with velocity - 2 I cap 2 JK per second to ICAP 2 JK m per second respectively the closes distance of approach and time when this happen respectively are? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two particle ab are point - 2 m - 2 m and 2 m - 2 m at t = 0 and moving with velocity - 2 I cap 2 JK per second to ICAP 2 JK m per second respectively the closes distance of approach and time when this happen respectively are?.
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