1The correct answer is 2 Ns
Two blocks A and B of masses 2 kg and 1 kg respectively are conected as shown and are at rest. The pulleys P1 and P2 are light and frictionless. A shell C of mass 1 kg moving vertically up with velocity 10 m/s collides with block B and gets permanently stuck to it.
Subsequently when the string just becomes taut the impulse developed in the string connected to B is???
Pls provide the detailed solution to this question....
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4 Answers
1I got 20Ns
velocity of B after collision can be found out by law of conservation of linear momentum.
→ 1 x 10 = (1+1)v'
→v' = 5m/s ( v' is velocity of B )
we need to find after wat time B comes to rest using v = u + at
0 = 5 - 10t → t = 0.5 sec
on reaching a height b comes to rest,
accln. experienced by B is 'g' downwards,
force on B = 2 x 10 = 20N,( 2kg since C stucks to B )
impulse for t = 0.5 sec, J = F.t →20 x 0.5 = 10 Ns
similarly force on A = 2 x 10 = 20N, Impulse J = 20 x 0.5 = 10 Ns
Net impulse is 10 + 10 Ns = 20 Ns
106velocity of B + ball combined = 5m/s upwards
Now, after string becomes loose, both A and B' (B+mass) will undergo free fall, as T=0 till the string becomes taut.
find the time when string becomes taut (by conserving length of string)
find the velocities of A and B'
Now use impulse momentum thm..This will give the answer