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A block of mass 10 kg start from rest and moves in a straight line under the effect of of a constant force of 10N. After 10 sec it collides head on with another block of mass 50 kg that was moving in the opposite direction at 2 m/s . If both the block stick together after the collision , what is the velocity of this combined entity ?
Most Upvoted Answer
A block of mass 10 kg start from rest and moves in a straight line und...
Calculation of Initial Velocity of the First Block
Calculation of Initial Velocity of the Second Block
Calculation of Velocity of Combined Entity After Collision
Conclusion
Therefore, the velocity of the combined entity after the collision is 1.33 m/s.
- Mass of the first block, m1 = 10 kg
- Force acting on the first block, F = 10 N
- Time taken, t = 10 sec
- Using the formula, F = ma, where a is the acceleration of the block
- Acceleration of the first block, a = F/m1 = 10/10 = 1 m/s²
- Using the formula, v = u + at, where u is the initial velocity of the block
- As the block was initially at rest, u = 0
- Velocity of the first block after 10 sec, v1 = u + at = 0 + 1 x 10 = 10 m/s
Calculation of Initial Velocity of the Second Block
- Mass of the second block, m2 = 50 kg
- Initial velocity of the second block, u2 = -2 m/s (as it was moving in the opposite direction)
Calculation of Velocity of Combined Entity After Collision
- Total momentum before the collision = m1u1 + m2u2
- Total momentum after the collision = (m1 + m2)v
- As the blocks stick together after the collision, their final velocity will be the same
- Using the law of conservation of momentum, m1u1 + m2u2 = (m1 + m2)v
- Substituting the values, 10 x 10 + 50 x (-2) = (10 + 50)v
- Solving the equation, v = 1.33 m/s
Conclusion
Therefore, the velocity of the combined entity after the collision is 1.33 m/s.
Community Answer
A block of mass 10 kg start from rest and moves in a straight line und...
The constant force produces a constant acceleration of, a = F/m = 10/10 = 1 m/sec sq
The velocity gained by the body in 10 sec is,
v = u + at = 0 + 1(10) = 10 m/s
Choosing the direction in which this body is moving to be positive we will now apply consent of momentum before and after the collision.
Mv + mv' = (M + m)v''
10(10) + 50(-2) = (10 + 50)v''
60v'' = 100 - 100
v'' = 0
The bodies will stick together and will stay at rest.
The velocity gained by the body in 10 sec is,
v = u + at = 0 + 1(10) = 10 m/s
Choosing the direction in which this body is moving to be positive we will now apply consent of momentum before and after the collision.
Mv + mv' = (M + m)v''
10(10) + 50(-2) = (10 + 50)v''
60v'' = 100 - 100
v'' = 0
The bodies will stick together and will stay at rest.
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A block of mass 10 kg start from rest and moves in a straight line under the effect of of a constant force of 10N. After 10 sec it collides head on with another block of mass 50 kg that was moving in the opposite direction at 2 m/s . If both the block stick together after the collision , what is the velocity of this combined entity ?
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A block of mass 10 kg start from rest and moves in a straight line under the effect of of a constant force of 10N. After 10 sec it collides head on with another block of mass 50 kg that was moving in the opposite direction at 2 m/s . If both the block stick together after the collision , what is the velocity of this combined entity ? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A block of mass 10 kg start from rest and moves in a straight line under the effect of of a constant force of 10N. After 10 sec it collides head on with another block of mass 50 kg that was moving in the opposite direction at 2 m/s . If both the block stick together after the collision , what is the velocity of this combined entity ? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 10 kg start from rest and moves in a straight line under the effect of of a constant force of 10N. After 10 sec it collides head on with another block of mass 50 kg that was moving in the opposite direction at 2 m/s . If both the block stick together after the collision , what is the velocity of this combined entity ?.
A block of mass 10 kg start from rest and moves in a straight line under the effect of of a constant force of 10N. After 10 sec it collides head on with another block of mass 50 kg that was moving in the opposite direction at 2 m/s . If both the block stick together after the collision , what is the velocity of this combined entity ? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A block of mass 10 kg start from rest and moves in a straight line under the effect of of a constant force of 10N. After 10 sec it collides head on with another block of mass 50 kg that was moving in the opposite direction at 2 m/s . If both the block stick together after the collision , what is the velocity of this combined entity ? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 10 kg start from rest and moves in a straight line under the effect of of a constant force of 10N. After 10 sec it collides head on with another block of mass 50 kg that was moving in the opposite direction at 2 m/s . If both the block stick together after the collision , what is the velocity of this combined entity ?.
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