Question
The blocks A and B shown in fig. Have masses MA=5kg and MB=4kg. The system is released from rest. The speed of B after A has travelled a distance 1 m along the incline is
Answer: Option C
:
C
If A moves down the incline by 1 m, B shal move up by 12m.
If the speed of B is v, then the speed of A will be 2v.
From conservation of energy,
Gain in KE = loss in PE
12mA(2v)2+12mBv2=mAg×35−mBg×12
Solving, we get
v=√g2√3
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:
C
If A moves down the incline by 1 m, B shal move up by 12m.
If the speed of B is v, then the speed of A will be 2v.
From conservation of energy,
Gain in KE = loss in PE
12mA(2v)2+12mBv2=mAg×35−mBg×12
Solving, we get
v=√g2√3
Was this answer helpful ?
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