Question

can someone please answer this question with explanation..........​

Answer 1

Just as the block B collides with C the block B will come to rest and the block C will move with a velocity of V towards right(because it is an elastic collision)

when two bodies of same mass one at rest and other moving collide, then one will stop and the other will start moving with same velocity.

now in the question the scenario will be something like shown figure just after collision and it will have the condition for max compression as shown in the figure below it:-

BY CONSERVATION OF MOMENTUM(Considering the spring block as a system(after collision)):-

2M×V=(2M+M)×V'

V'=⅔V. ...................(1)

BY CONSERVATION OF MECHANICAL ENERGY:-

$\frac{1}{2} \times 2m \times {v}^{2} = \frac{1}{2} \times (2m + m) \times {v(final)}^{2} + \frac{1}{2} \times k {x}^{2}$

now simply put all the values and solve the equation:-

$\frac{1}{2} 2m {v}^{2} = \frac{1}{2} \times 3m( { \frac{2v}{3} })^{2} + \frac{1}{2} \times k {x}^{2}$

$2m { {v}^{2}} = \frac{4}{3} \times m {v}^{2} + k {x}^{2}$

$(2 - \frac{4}{3} ) m {v}^{2} = k {x}^{2}$

$\frac{2m {v}^{2} }{3k} = {x}^{2}$

$\sqrt{ \frac{2m}{3k} } v = x$

write in comments for any doubts.

FOLLOW IF THAT HELPED.

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