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A block of mass 4 kg moving with a velocity of 10 m/s collides elastically with a block of mass 5 kg which is at rest.If after locks deflected at angle 30 degree and 60 degree respectively,then find the velocities of the block after collision. .Ans = 5√3 and 4 m/s.
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A block of mass 4 kg moving with a velocity of 10 m/s collides elastic...
Velocity of the first block before collision (u1) = 10 m/s
Mass of the first block (m1) = 4 kg
Velocity of the second block before collision (u2) = 0 m/s (at rest)
Mass of the second block (m2) = 5 kg
The collision between the two blocks is elastic, which means both momentum and kinetic energy are conserved.
1. Conservation of momentum:
Before collision: m1u1 + m2u2 = m1v1 + m2v2
After collision, the two blocks move in different directions. Let's take the positive direction as the direction in which the first block moves.
m1u1 - m2u2 = m1v1 - m2v2 (Equation 1)
2. Conservation of kinetic energy:
Before collision: (1/2)m1u1^2 + (1/2)m2u2^2 = (1/2)m1v1^2 + (1/2)m2v2^2
Since the collision is elastic, the kinetic energy is conserved.
(1/2)m1u1^2 = (1/2)m1v1^2 + (1/2)m2v2^2 (Equation 2)
3. Solving for the velocities:
Using equation 1, we can solve for v1:
m1u1 - m2u2 = m1v1 - m2v2
4(10) - 5(0) = 4v1 - 5v2
40 = 4v1 - 5v2 (Equation 3)
Using equation 2, we can solve for v2:
(1/2)m1u1^2 = (1/2)m1v1^2 + (1/2)m2v2^2
2(100) = 4v1^2 + 5v2^2
200 = 4v1^2 + 5v2^2 (Equation 4)
4. Solving the equations:
Multiplying equation 3 by 4 and equation 4 by 5, we can eliminate the v2 term:
160 = 16v1 - 20v2
1000 = 20v1^2 + 25v2^2
Multiplying equation 3 by 5 and equation 4 by 4, we can eliminate the v1 term:
200 = 20v1 - 25v2
800 = 16v1^2 + 20v2^2
Adding the two resulting equations:
160 + 200 = 16v1 - 20v2 + 20v1 - 25v2
360 = 36v1 - 45v2
Rearranging the equation:
36v1 - 45v2 = 360
Using the angle of deflection, we can express v1 and v2 in terms of trigonometric functions:
v1 = v cos(30)
v2 = v sin(60)
Substituting these values into the equation:
36v cos(30) - 45v sin(60) = 360
Simplifying:
36(√3/2)v - 45(√3/2)v
Mass of the first block (m1) = 4 kg
Velocity of the second block before collision (u2) = 0 m/s (at rest)
Mass of the second block (m2) = 5 kg
The collision between the two blocks is elastic, which means both momentum and kinetic energy are conserved.
1. Conservation of momentum:
Before collision: m1u1 + m2u2 = m1v1 + m2v2
After collision, the two blocks move in different directions. Let's take the positive direction as the direction in which the first block moves.
m1u1 - m2u2 = m1v1 - m2v2 (Equation 1)
2. Conservation of kinetic energy:
Before collision: (1/2)m1u1^2 + (1/2)m2u2^2 = (1/2)m1v1^2 + (1/2)m2v2^2
Since the collision is elastic, the kinetic energy is conserved.
(1/2)m1u1^2 = (1/2)m1v1^2 + (1/2)m2v2^2 (Equation 2)
3. Solving for the velocities:
Using equation 1, we can solve for v1:
m1u1 - m2u2 = m1v1 - m2v2
4(10) - 5(0) = 4v1 - 5v2
40 = 4v1 - 5v2 (Equation 3)
Using equation 2, we can solve for v2:
(1/2)m1u1^2 = (1/2)m1v1^2 + (1/2)m2v2^2
2(100) = 4v1^2 + 5v2^2
200 = 4v1^2 + 5v2^2 (Equation 4)
4. Solving the equations:
Multiplying equation 3 by 4 and equation 4 by 5, we can eliminate the v2 term:
160 = 16v1 - 20v2
1000 = 20v1^2 + 25v2^2
Multiplying equation 3 by 5 and equation 4 by 4, we can eliminate the v1 term:
200 = 20v1 - 25v2
800 = 16v1^2 + 20v2^2
Adding the two resulting equations:
160 + 200 = 16v1 - 20v2 + 20v1 - 25v2
360 = 36v1 - 45v2
Rearranging the equation:
36v1 - 45v2 = 360
Using the angle of deflection, we can express v1 and v2 in terms of trigonometric functions:
v1 = v cos(30)
v2 = v sin(60)
Substituting these values into the equation:
36v cos(30) - 45v sin(60) = 360
Simplifying:
36(√3/2)v - 45(√3/2)v
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A block of mass 4 kg moving with a velocity of 10 m/s collides elastically with a block of mass 5 kg which is at rest.If after locks deflected at angle 30 degree and 60 degree respectively,then find the velocities of the block after collision. .Ans = 5√3 and 4 m/s.
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A block of mass 4 kg moving with a velocity of 10 m/s collides elastically with a block of mass 5 kg which is at rest.If after locks deflected at angle 30 degree and 60 degree respectively,then find the velocities of the block after collision. .Ans = 5√3 and 4 m/s. for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A block of mass 4 kg moving with a velocity of 10 m/s collides elastically with a block of mass 5 kg which is at rest.If after locks deflected at angle 30 degree and 60 degree respectively,then find the velocities of the block after collision. .Ans = 5√3 and 4 m/s. covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 4 kg moving with a velocity of 10 m/s collides elastically with a block of mass 5 kg which is at rest.If after locks deflected at angle 30 degree and 60 degree respectively,then find the velocities of the block after collision. .Ans = 5√3 and 4 m/s..
A block of mass 4 kg moving with a velocity of 10 m/s collides elastically with a block of mass 5 kg which is at rest.If after locks deflected at angle 30 degree and 60 degree respectively,then find the velocities of the block after collision. .Ans = 5√3 and 4 m/s. for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A block of mass 4 kg moving with a velocity of 10 m/s collides elastically with a block of mass 5 kg which is at rest.If after locks deflected at angle 30 degree and 60 degree respectively,then find the velocities of the block after collision. .Ans = 5√3 and 4 m/s. covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 4 kg moving with a velocity of 10 m/s collides elastically with a block of mass 5 kg which is at rest.If after locks deflected at angle 30 degree and 60 degree respectively,then find the velocities of the block after collision. .Ans = 5√3 and 4 m/s..
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A block of mass 4 kg moving with a velocity of 10 m/s collides elastically with a block of mass 5 kg which is at rest.If after locks deflected at angle 30 degree and 60 degree respectively,then find the velocities of the block after collision. .Ans = 5√3 and 4 m/s., a detailed solution for A block of mass 4 kg moving with a velocity of 10 m/s collides elastically with a block of mass 5 kg which is at rest.If after locks deflected at angle 30 degree and 60 degree respectively,then find the velocities of the block after collision. .Ans = 5√3 and 4 m/s. has been provided alongside types of A block of mass 4 kg moving with a velocity of 10 m/s collides elastically with a block of mass 5 kg which is at rest.If after locks deflected at angle 30 degree and 60 degree respectively,then find the velocities of the block after collision. .Ans = 5√3 and 4 m/s. theory, EduRev gives you an
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