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A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be
- a)0.5
- b)0.25
- c)0.8
- d)0.4
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A moving block having mass m, collides with another stationary block ...
Given:
- Mass of the moving block = m
- Mass of the stationary block = 4m
- Initial velocity of the lighter block = v
- Coefficient of restitution = e
Concepts:
- Coefficient of restitution (e) is a measure of the elasticity of a collision between two objects. It is defined as the ratio of the relative velocity of separation to the relative velocity of approach after a collision.
- The formula for coefficient of restitution is given by: e = (v2 - v1) / (u1 - u2), where v1 and v2 are the final velocities of the two objects and u1 and u2 are their initial velocities.
Analysis:
- Let the final velocity of the lighter block be v1' and the final velocity of the heavier block be v2'.
- Since the lighter block comes to rest after the collision, v1' = 0.
- The initial velocity of the heavier block is 0 (as it is stationary).
- Using the formula for coefficient of restitution, we can write: e = (v2' - 0) / (v - 0) = v2' / v.
Solution:
- We need to find the value of coefficient of restitution (e).
- From the given information, we know that the lighter block comes to rest after the collision, i.e., v1' = 0.
- Since the initial velocity of the heavier block is 0, v2' is also 0.
- Therefore, e = v2' / v = 0 / v = 0.
Answer:
The value of coefficient of restitution (e) is 0.
Explanation:
- The coefficient of restitution measures the elasticity of a collision. A value of 0 means the collision is perfectly inelastic, i.e., the objects stick together after the collision and move with a common velocity.
- In this case, since the lighter block comes to rest after the collision, it indicates a perfectly inelastic collision.
- Therefore, the value of coefficient of restitution is 0, which is given in option B.
- Mass of the moving block = m
- Mass of the stationary block = 4m
- Initial velocity of the lighter block = v
- Coefficient of restitution = e
Concepts:
- Coefficient of restitution (e) is a measure of the elasticity of a collision between two objects. It is defined as the ratio of the relative velocity of separation to the relative velocity of approach after a collision.
- The formula for coefficient of restitution is given by: e = (v2 - v1) / (u1 - u2), where v1 and v2 are the final velocities of the two objects and u1 and u2 are their initial velocities.
Analysis:
- Let the final velocity of the lighter block be v1' and the final velocity of the heavier block be v2'.
- Since the lighter block comes to rest after the collision, v1' = 0.
- The initial velocity of the heavier block is 0 (as it is stationary).
- Using the formula for coefficient of restitution, we can write: e = (v2' - 0) / (v - 0) = v2' / v.
Solution:
- We need to find the value of coefficient of restitution (e).
- From the given information, we know that the lighter block comes to rest after the collision, i.e., v1' = 0.
- Since the initial velocity of the heavier block is 0, v2' is also 0.
- Therefore, e = v2' / v = 0 / v = 0.
Answer:
The value of coefficient of restitution (e) is 0.
Explanation:
- The coefficient of restitution measures the elasticity of a collision. A value of 0 means the collision is perfectly inelastic, i.e., the objects stick together after the collision and move with a common velocity.
- In this case, since the lighter block comes to rest after the collision, it indicates a perfectly inelastic collision.
- Therefore, the value of coefficient of restitution is 0, which is given in option B.
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Community Answer
A moving block having mass m, collides with another stationary block ...
Coefficient of Restitution (e) -
Ratio of relative velocity after collision to relative velocity before collision.
Let velocity of 4m mass is v after collision
then mv = 4mv
v = v/4v
= 1/4
= 0.25
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A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will bea)0.5b)0.25c)0.8d)0.4Correct answer is option 'B'. Can you explain this answer?
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A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will bea)0.5b)0.25c)0.8d)0.4Correct answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will bea)0.5b)0.25c)0.8d)0.4Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will bea)0.5b)0.25c)0.8d)0.4Correct answer is option 'B'. Can you explain this answer?.
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