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Two blocks of masses 5 kg and 2 kg are placed on a frictionless surface and connected by a spring. an external kick gives a velocity of 14 m/s to the heavier block in the direction of lighter one.the velocity gained by the center of mass is?
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Two blocks of masses 5 kg and 2 kg are placed on a frictionless surfac...
Momentum of first ball = Momentum of the second ball
m1 v1 = m2 v2
5 x 14 = 2 x v2
v2 = 35 m/s
Velocity of centre of mass:
v(cm) = [m1v1 + m2v2]/m1+m2
= [(5x14)+(2x35)]/(5+2)
= 20 m/s
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Two blocks of masses 5 kg and 2 kg are placed on a frictionless surfac...
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Two blocks of masses 5 kg and 2 kg are placed on a frictionless surfac...
Given information:
- Mass of the first block (heavier block) = 5 kg
- Mass of the second block (lighter block) = 2 kg
- Velocity of the first block after the kick = 14 m/s
- The direction of the velocity is towards the second block
Calculation of the velocity gained by the center of mass:
Step 1: Calculate the total mass of the system:
The total mass of the system is the sum of the masses of both blocks.
Total mass = Mass of the first block + Mass of the second block
Total mass = 5 kg + 2 kg = 7 kg
Step 2: Calculate the velocity of the center of mass:
The velocity of the center of mass can be calculated using the principle of conservation of momentum. According to this principle, the total momentum before the kick is equal to the total momentum after the kick.
Total momentum before the kick = Total momentum after the kick
Momentum is calculated as the product of mass and velocity.
Total momentum before the kick = (Mass of the first block × Velocity of the first block) + (Mass of the second block × Velocity of the second block)
The velocity of the second block is unknown, but we can solve for it.
Momentum before the kick = Momentum after the kick
(Mass of the first block × Velocity of the first block) + (Mass of the second block × Velocity of the second block) = Total mass × Velocity of the center of mass
Substituting the known values:
(5 kg × 14 m/s) + (2 kg × Velocity of the second block) = 7 kg × Velocity of the center of mass
70 kg·m/s + 2 kg × Velocity of the second block = 7 kg × Velocity of the center of mass
Step 3: Solve for the velocity of the center of mass:
Rearrange the equation to solve for the velocity of the center of mass.
Velocity of the center of mass = (70 kg·m/s + 2 kg × Velocity of the second block) / 7 kg
Step 4: Substitute the given values:
Velocity of the center of mass = (70 kg·m/s + 2 kg × Velocity of the second block) / 7 kg
Now, since the blocks are connected by a spring and there is no external force acting on the system after the kick, the total momentum of the system remains constant. Therefore, the velocity of the center of mass remains constant at the value calculated in step 3.
Conclusion:
The velocity gained by the center of mass of the system is calculated to be (70 kg·m/s + 2 kg × Velocity of the second block) / 7 kg. This value represents the average velocity of the entire system.
- Mass of the first block (heavier block) = 5 kg
- Mass of the second block (lighter block) = 2 kg
- Velocity of the first block after the kick = 14 m/s
- The direction of the velocity is towards the second block
Calculation of the velocity gained by the center of mass:
Step 1: Calculate the total mass of the system:
The total mass of the system is the sum of the masses of both blocks.
Total mass = Mass of the first block + Mass of the second block
Total mass = 5 kg + 2 kg = 7 kg
Step 2: Calculate the velocity of the center of mass:
The velocity of the center of mass can be calculated using the principle of conservation of momentum. According to this principle, the total momentum before the kick is equal to the total momentum after the kick.
Total momentum before the kick = Total momentum after the kick
Momentum is calculated as the product of mass and velocity.
Total momentum before the kick = (Mass of the first block × Velocity of the first block) + (Mass of the second block × Velocity of the second block)
The velocity of the second block is unknown, but we can solve for it.
Momentum before the kick = Momentum after the kick
(Mass of the first block × Velocity of the first block) + (Mass of the second block × Velocity of the second block) = Total mass × Velocity of the center of mass
Substituting the known values:
(5 kg × 14 m/s) + (2 kg × Velocity of the second block) = 7 kg × Velocity of the center of mass
70 kg·m/s + 2 kg × Velocity of the second block = 7 kg × Velocity of the center of mass
Step 3: Solve for the velocity of the center of mass:
Rearrange the equation to solve for the velocity of the center of mass.
Velocity of the center of mass = (70 kg·m/s + 2 kg × Velocity of the second block) / 7 kg
Step 4: Substitute the given values:
Velocity of the center of mass = (70 kg·m/s + 2 kg × Velocity of the second block) / 7 kg
Now, since the blocks are connected by a spring and there is no external force acting on the system after the kick, the total momentum of the system remains constant. Therefore, the velocity of the center of mass remains constant at the value calculated in step 3.
Conclusion:
The velocity gained by the center of mass of the system is calculated to be (70 kg·m/s + 2 kg × Velocity of the second block) / 7 kg. This value represents the average velocity of the entire system.
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Two blocks of masses 5 kg and 2 kg are placed on a frictionless surface and connected by a spring. an external kick gives a velocity of 14 m/s to the heavier block in the direction of lighter one.the velocity gained by the center of mass is?
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Two blocks of masses 5 kg and 2 kg are placed on a frictionless surface and connected by a spring. an external kick gives a velocity of 14 m/s to the heavier block in the direction of lighter one.the velocity gained by the center of mass is? 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 blocks of masses 5 kg and 2 kg are placed on a frictionless surface and connected by a spring. an external kick gives a velocity of 14 m/s to the heavier block in the direction of lighter one.the velocity gained by the center of mass is? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two blocks of masses 5 kg and 2 kg are placed on a frictionless surface and connected by a spring. an external kick gives a velocity of 14 m/s to the heavier block in the direction of lighter one.the velocity gained by the center of mass is?.
Two blocks of masses 5 kg and 2 kg are placed on a frictionless surface and connected by a spring. an external kick gives a velocity of 14 m/s to the heavier block in the direction of lighter one.the velocity gained by the center of mass is? 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 blocks of masses 5 kg and 2 kg are placed on a frictionless surface and connected by a spring. an external kick gives a velocity of 14 m/s to the heavier block in the direction of lighter one.the velocity gained by the center of mass is? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two blocks of masses 5 kg and 2 kg are placed on a frictionless surface and connected by a spring. an external kick gives a velocity of 14 m/s to the heavier block in the direction of lighter one.the velocity gained by the center of mass is?.
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