JEE Exam > JEE Questions > Figure shows block A of mass 0.2 kg sliding t...
Start Learning for Free
Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed spring of spring constant 1000 Nm–1. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance 'd' from the base of that surface after falling height 5m. (use π2 = 10; g = 10 m/s2) Assume that the spring does not affect the collision.
Amplitude of the SHM as being executed by block B-spring system, is -
- a)2.5√10 cm
- b)10 cm
- c)3√10cm
- d)5√10cm
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
Figure shows block A of mass 0.2 kg sliding to the right over a fricti...
Immediately after the collision, suppose velocities of the blocks are V1 and V1 as shown 1/2 vel. of approach = velocity of separation.
⇒ 5 = V2 - V1 .... (1)
Using principle of conservation of momentum for the collision
2 = 0.2 V1 + V2
or 10 = V1 + 5V2 .....(2)
On solving V2 = 2.5 m/s; V1 = - 2.5 m/s
Hence block A moves leftward after the collision with speed 2.5 m/s. And the block B moves towards right with speed 2.5 m/s.
The maximum velocity of B = 2.5 = ωA
⇒
⇒ 5 = V2 - V1 .... (1)
Using principle of conservation of momentum for the collision
2 = 0.2 V1 + V2
or 10 = V1 + 5V2 .....(2)
On solving V2 = 2.5 m/s; V1 = - 2.5 m/s
Hence block A moves leftward after the collision with speed 2.5 m/s. And the block B moves towards right with speed 2.5 m/s.
The maximum velocity of B = 2.5 = ωA
⇒
Most Upvoted Answer
Figure shows block A of mass 0.2 kg sliding to the right over a fricti...
Immediately after the collision, suppose velocities of the blocks are V1 and V1 as shown 1/2 vel. of approach = velocity of separation.
⇒ 5 = V2 - V1 .... (1)
Using principle of conservation of momentum for the collision
2 = 0.2 V1 + V2
or 10 = V1 + 5V2 .....(2)
On solving V2 = 2.5 m/s; V1 = - 2.5 m/s
Hence block A moves leftward after the collision with speed 2.5 m/s. And the block B moves towards right with speed 2.5 m/s.
The maximum velocity of B = 2.5 = ωA
⇒
⇒ 5 = V2 - V1 .... (1)
Using principle of conservation of momentum for the collision
2 = 0.2 V1 + V2
or 10 = V1 + 5V2 .....(2)
On solving V2 = 2.5 m/s; V1 = - 2.5 m/s
Hence block A moves leftward after the collision with speed 2.5 m/s. And the block B moves towards right with speed 2.5 m/s.
The maximum velocity of B = 2.5 = ωA
⇒
Free Test
FREE
| Start Free Test |
Community Answer
Figure shows block A of mass 0.2 kg sliding to the right over a fricti...
Immediately after the collision, suppose velocities of the blocks are V1 and V1 as shown 1/2 vel. of approach = velocity of separation.
⇒ 5 = V2 - V1 .... (1)
Using principle of conservation of momentum for the collision
2 = 0.2 V1 + V2
or 10 = V1 + 5V2 .....(2)
On solving V2 = 2.5 m/s; V1 = - 2.5 m/s
Hence block A moves leftward after the collision with speed 2.5 m/s. And the block B moves towards right with speed 2.5 m/s.
The maximum velocity of B = 2.5 = ωA
⇒
⇒ 5 = V2 - V1 .... (1)
Using principle of conservation of momentum for the collision
2 = 0.2 V1 + V2
or 10 = V1 + 5V2 .....(2)
On solving V2 = 2.5 m/s; V1 = - 2.5 m/s
Hence block A moves leftward after the collision with speed 2.5 m/s. And the block B moves towards right with speed 2.5 m/s.
The maximum velocity of B = 2.5 = ωA
⇒
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
Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed springof spring constant 1000 Nm–1. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance d from the base of that surface after falling height 5m.(use π2 = 10; g = 10 m/s2) Assume that the springdoes not affect the collision.Amplitude of the SHM as being executed by block B-spring system, is -a)2.5√10 cmb)10 cmc)3√10cmd)5√10cmCorrect answer is option 'A'. Can you explain this answer?
Question Description
Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed springof spring constant 1000 Nm–1. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance d from the base of that surface after falling height 5m.(use π2 = 10; g = 10 m/s2) Assume that the springdoes not affect the collision.Amplitude of the SHM as being executed by block B-spring system, is -a)2.5√10 cmb)10 cmc)3√10cmd)5√10cmCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed springof spring constant 1000 Nm–1. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance d from the base of that surface after falling height 5m.(use π2 = 10; g = 10 m/s2) Assume that the springdoes not affect the collision.Amplitude of the SHM as being executed by block B-spring system, is -a)2.5√10 cmb)10 cmc)3√10cmd)5√10cmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed springof spring constant 1000 Nm–1. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance d from the base of that surface after falling height 5m.(use π2 = 10; g = 10 m/s2) Assume that the springdoes not affect the collision.Amplitude of the SHM as being executed by block B-spring system, is -a)2.5√10 cmb)10 cmc)3√10cmd)5√10cmCorrect answer is option 'A'. Can you explain this answer?.
Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed springof spring constant 1000 Nm–1. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance d from the base of that surface after falling height 5m.(use π2 = 10; g = 10 m/s2) Assume that the springdoes not affect the collision.Amplitude of the SHM as being executed by block B-spring system, is -a)2.5√10 cmb)10 cmc)3√10cmd)5√10cmCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed springof spring constant 1000 Nm–1. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance d from the base of that surface after falling height 5m.(use π2 = 10; g = 10 m/s2) Assume that the springdoes not affect the collision.Amplitude of the SHM as being executed by block B-spring system, is -a)2.5√10 cmb)10 cmc)3√10cmd)5√10cmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed springof spring constant 1000 Nm–1. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance d from the base of that surface after falling height 5m.(use π2 = 10; g = 10 m/s2) Assume that the springdoes not affect the collision.Amplitude of the SHM as being executed by block B-spring system, is -a)2.5√10 cmb)10 cmc)3√10cmd)5√10cmCorrect answer is option 'A'. Can you explain this answer?.
Solutions for Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed springof spring constant 1000 Nm–1. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance d from the base of that surface after falling height 5m.(use π2 = 10; g = 10 m/s2) Assume that the springdoes not affect the collision.Amplitude of the SHM as being executed by block B-spring system, is -a)2.5√10 cmb)10 cmc)3√10cmd)5√10cmCorrect answer is option 'A'. Can you explain this answer? 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 Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed springof spring constant 1000 Nm–1. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance d from the base of that surface after falling height 5m.(use π2 = 10; g = 10 m/s2) Assume that the springdoes not affect the collision.Amplitude of the SHM as being executed by block B-spring system, is -a)2.5√10 cmb)10 cmc)3√10cmd)5√10cmCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed springof spring constant 1000 Nm–1. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance d from the base of that surface after falling height 5m.(use π2 = 10; g = 10 m/s2) Assume that the springdoes not affect the collision.Amplitude of the SHM as being executed by block B-spring system, is -a)2.5√10 cmb)10 cmc)3√10cmd)5√10cmCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed springof spring constant 1000 Nm–1. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance d from the base of that surface after falling height 5m.(use π2 = 10; g = 10 m/s2) Assume that the springdoes not affect the collision.Amplitude of the SHM as being executed by block B-spring system, is -a)2.5√10 cmb)10 cmc)3√10cmd)5√10cmCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed springof spring constant 1000 Nm–1. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance d from the base of that surface after falling height 5m.(use π2 = 10; g = 10 m/s2) Assume that the springdoes not affect the collision.Amplitude of the SHM as being executed by block B-spring system, is -a)2.5√10 cmb)10 cmc)3√10cmd)5√10cmCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed springof spring constant 1000 Nm–1. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance d from the base of that surface after falling height 5m.(use π2 = 10; g = 10 m/s2) Assume that the springdoes not affect the collision.Amplitude of the SHM as being executed by block B-spring system, is -a)2.5√10 cmb)10 cmc)3√10cmd)5√10cmCorrect answer is option 'A'. Can you explain this answer? 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
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup