JEE Exam > JEE Questions > A piece of wood of mass 0.03 kg is dropped fr...
Start Learning for Free
A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms–1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is : (g =10ms–2) (1) 30 m (2) 10 m (3) 40 m (4) 20 m?
Verified Answer
A piece of wood of mass 0.03 kg is dropped from the top of a 100 m hei...
Ans.
initially at t = 0 COM was at a height of 60 m from ground.
∴
Height reached by COM (which is same as the position of blocks after collision). Will be = 60 + 80 = 140 cm from ground.
∴
40 m
This question is part of UPSC exam. View all JEE courses
Most Upvoted Answer
A piece of wood of mass 0.03 kg is dropped from the top of a 100 m hei...
Given information:
- Mass of wood, m1 = 0.03 kg
- Mass of bullet, m2 = 0.02 kg
- Height of the building, h = 100 m
- Initial velocity of the bullet, u = 100 m/s
- Acceleration due to gravity, g = 10 m/s²
Analysis:
- When the wood is dropped from the top of the building, it falls freely under the influence of gravity.
- The bullet is fired vertically upward from the ground, opposing the force of gravity.
- The bullet and wood collide and stick together, forming a single system.
- The combined system rises to a maximum height before falling back down due to the force of gravity.
Solution:
To find the maximum height reached by the combined system, we need to analyze the conservation of energy.
Step 1: Determine the initial and final kinetic energies:
- The initial kinetic energy of the wood is zero as it is dropped from rest.
- The initial kinetic energy of the bullet is given by:
KE_initial = (1/2) * m2 * u²
Step 2: Determine the potential energy at the maximum height:
- At the maximum height, the velocity of the combined system is zero.
- The potential energy at the maximum height is given by:
PE_max = (m1 + m2) * g * h_max
Step 3: Equate the initial and final energies:
- According to the law of conservation of energy, the initial kinetic energy of the system should be equal to the potential energy at the maximum height.
- So, we equate the initial kinetic energy with the potential energy at the maximum height:
(1/2) * m2 * u² = (m1 + m2) * g * h_max
Step 4: Solve for the maximum height:
- Substituting the given values, we have:
(1/2) * 0.02 * (100)² = (0.03 + 0.02) * 10 * h_max
1 = 0.05 * h_max
h_max = 20 m
Answer:
The maximum height to which the combined system reaches above the top of the building before falling below is 20 m. (Option 4)
- Mass of wood, m1 = 0.03 kg
- Mass of bullet, m2 = 0.02 kg
- Height of the building, h = 100 m
- Initial velocity of the bullet, u = 100 m/s
- Acceleration due to gravity, g = 10 m/s²
Analysis:
- When the wood is dropped from the top of the building, it falls freely under the influence of gravity.
- The bullet is fired vertically upward from the ground, opposing the force of gravity.
- The bullet and wood collide and stick together, forming a single system.
- The combined system rises to a maximum height before falling back down due to the force of gravity.
Solution:
To find the maximum height reached by the combined system, we need to analyze the conservation of energy.
Step 1: Determine the initial and final kinetic energies:
- The initial kinetic energy of the wood is zero as it is dropped from rest.
- The initial kinetic energy of the bullet is given by:
KE_initial = (1/2) * m2 * u²
Step 2: Determine the potential energy at the maximum height:
- At the maximum height, the velocity of the combined system is zero.
- The potential energy at the maximum height is given by:
PE_max = (m1 + m2) * g * h_max
Step 3: Equate the initial and final energies:
- According to the law of conservation of energy, the initial kinetic energy of the system should be equal to the potential energy at the maximum height.
- So, we equate the initial kinetic energy with the potential energy at the maximum height:
(1/2) * m2 * u² = (m1 + m2) * g * h_max
Step 4: Solve for the maximum height:
- Substituting the given values, we have:
(1/2) * 0.02 * (100)² = (0.03 + 0.02) * 10 * h_max
1 = 0.05 * h_max
h_max = 20 m
Answer:
The maximum height to which the combined system reaches above the top of the building before falling below is 20 m. (Option 4)
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms–1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is : (g =10ms–2) (1) 30 m (2) 10 m (3) 40 m (4) 20 m?
Question Description
A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms–1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is : (g =10ms–2) (1) 30 m (2) 10 m (3) 40 m (4) 20 m? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms–1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is : (g =10ms–2) (1) 30 m (2) 10 m (3) 40 m (4) 20 m? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms–1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is : (g =10ms–2) (1) 30 m (2) 10 m (3) 40 m (4) 20 m?.
A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms–1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is : (g =10ms–2) (1) 30 m (2) 10 m (3) 40 m (4) 20 m? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms–1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is : (g =10ms–2) (1) 30 m (2) 10 m (3) 40 m (4) 20 m? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms–1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is : (g =10ms–2) (1) 30 m (2) 10 m (3) 40 m (4) 20 m?.
Solutions for A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms–1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is : (g =10ms–2) (1) 30 m (2) 10 m (3) 40 m (4) 20 m? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms–1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is : (g =10ms–2) (1) 30 m (2) 10 m (3) 40 m (4) 20 m? defined & explained in the simplest way possible. Besides giving the explanation of
A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms–1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is : (g =10ms–2) (1) 30 m (2) 10 m (3) 40 m (4) 20 m?, a detailed solution for A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms–1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is : (g =10ms–2) (1) 30 m (2) 10 m (3) 40 m (4) 20 m? has been provided alongside types of A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms–1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is : (g =10ms–2) (1) 30 m (2) 10 m (3) 40 m (4) 20 m? theory, EduRev gives you an
ample number of questions to practice A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms–1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is : (g =10ms–2) (1) 30 m (2) 10 m (3) 40 m (4) 20 m? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup