Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
Two blocks A and B each of mass m are connected by massless spring of natural length L and spring constant k. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length as shown in figure. A third identical block C also of mass m moves on the floor with a speed v along the line joining A and B and collides elastically with A, then
(1) the maximum compression of the Spring is \[v\sqrt{m/k}\] (2) the maximum compression of the spring is \[v\sqrt{m/2k}\] (3) the kinetic energy of the A-B system at maximum compression of the spring is zero (4) the kinetic energy of the A-B system at maximum compression of the spring is\[\frac{m{{v}^{2}}}{4}\]
A) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: C
Solution :
Initially A and B are at rest on smooth surface. Then C moves with velocity v, collides elastically with A and stops. As a result of head-on collision between C and A block. A acquires a velocity v and moves towards B. It compresses spring L which pushes B towards right A goes on compressing the spring till the velocity acquired by B becomes equal to the velocity of A. Let this velocity be\[v'\]. This state occurs when the spring is in a state of maximum compression. Let\[x\]be the maximum compression in spring at this stage. By conservation of linear mementum, we get or \[mv=mv'+mv'\]or\[v'=\frac{v}{2}\] ...(i) By conservation of mechanical energy, we get \[\therefore \] \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}m{{v}^{2}}+\frac{1}{2}k{{x}^{2}}\] Or\[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{\left( \frac{v}{2} \right)}^{2}}+\frac{1}{2}m{{\left( \frac{v}{2} \right)}^{2}}+\frac{1}{2}k{{x}^{2}}\] \[\therefore \]\[\frac{1}{2}m{{v}^{2}}=\frac{1}{4}m{{v}^{2}}+\frac{1}{2}k{{x}^{2}}\]or\[\frac{k{{x}^{2}}}{2}=\frac{m{{v}^{2}}}{4}\] Or \[x=v\sqrt{\frac{m}{2k}}\] At maximum compression of the spring, the kinetic energy of A-B system is \[=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}mv{{'}^{2}}=mv{{'}^{2}}=m{{\left( \frac{v}{2} \right)}^{2}}=\frac{m{{v}^{2}}}{4}\]
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