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Two block of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is
[IIT(Scr.)-2002]
- a)30 m/s
- b)20 m/s
- c)10 m/s
- d)5 m/s
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Two block of masses 10 kg and 4 kg are connected by a spring of neglig...
Velocity of heavier block after the impulse = 14m/s
Velocity of lighter block after the impulse = 0 m/s
Thus velocity of COM = 14 x 10 + 0 x 4 / 10 + 4 = 10m/s
Velocity of lighter block after the impulse = 0 m/s
Thus velocity of COM = 14 x 10 + 0 x 4 / 10 + 4 = 10m/s
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Two block of masses 10 kg and 4 kg are connected by a spring of neglig...
Understanding the Problem
In this scenario, we have two blocks with masses of 10 kg and 4 kg connected by a spring on a frictionless surface. An impulse is applied to the 10 kg block, giving it a velocity of 14 m/s towards the lighter block.
Calculating Total Mass
- Mass of Block 1 (m1) = 10 kg
- Mass of Block 2 (m2) = 4 kg
- Total mass (M) = m1 + m2 = 10 kg + 4 kg = 14 kg
Impulse and Velocity of the Heavier Block
- The impulse gives the heavier block (10 kg) a velocity of 14 m/s.
- This velocity is directed toward the 4 kg block.
Calculating Center of Mass Velocity
To find the velocity of the center of mass (V_cm), we can use the formula:
V_cm = (m1 * v1 + m2 * v2) / (m1 + m2)
Where:
- v1 = velocity of the heavier block = 14 m/s
- v2 = velocity of the lighter block = 0 m/s (initially at rest)
Substituting the values:
V_cm = [(10 kg * 14 m/s) + (4 kg * 0 m/s)] / (14 kg)
V_cm = (140 kg*m/s) / (14 kg)
Final Result
- V_cm = 10 m/s
Thus, the velocity of the center of mass is 10 m/s, making the correct answer option (c).
This explains why option 'C' is the correct answer to the problem.
In this scenario, we have two blocks with masses of 10 kg and 4 kg connected by a spring on a frictionless surface. An impulse is applied to the 10 kg block, giving it a velocity of 14 m/s towards the lighter block.
Calculating Total Mass
- Mass of Block 1 (m1) = 10 kg
- Mass of Block 2 (m2) = 4 kg
- Total mass (M) = m1 + m2 = 10 kg + 4 kg = 14 kg
Impulse and Velocity of the Heavier Block
- The impulse gives the heavier block (10 kg) a velocity of 14 m/s.
- This velocity is directed toward the 4 kg block.
Calculating Center of Mass Velocity
To find the velocity of the center of mass (V_cm), we can use the formula:
V_cm = (m1 * v1 + m2 * v2) / (m1 + m2)
Where:
- v1 = velocity of the heavier block = 14 m/s
- v2 = velocity of the lighter block = 0 m/s (initially at rest)
Substituting the values:
V_cm = [(10 kg * 14 m/s) + (4 kg * 0 m/s)] / (14 kg)
V_cm = (140 kg*m/s) / (14 kg)
Final Result
- V_cm = 10 m/s
Thus, the velocity of the center of mass is 10 m/s, making the correct answer option (c).
This explains why option 'C' is the correct answer to the problem.
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Two block of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is [IIT(Scr.)-2002]a)30 m/sb)20 m/sc)10 m/sd)5 m/sCorrect answer is option 'C'. Can you explain this answer?
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