Question

pls solve this Q.....!!!! ​

Answer 1

Answer:

okay.....here is ur answer..........

Answer 2

Here, Momentum will remain same before the collision as well as after the collision.

After collision both the blocks And B, will move with same velocity for some instant.

At that instant let the combined velocity be V

Now,

$= > Initial \: momentum=2 \times 1=2 \\ \\ = > Final \: momentum=(m+m) \times v=2v \\ = > 2=2v \\ = > v=1m/s$

Fom work energy theorem we know that change in kinetic energy is total work done.

$= > W_{all}= \Delta K$

$Change \: in \: kinetic \: energy= \frac{1}{2} \times 2 \times {1}^{2} - \frac{1}{2} \times 1 \times 4 \\ =1-2 \\ = 1$

(KE is always taken as positive)

Work done = 1

$Total \: work \: done= \frac{1}{2} k {x}^{2}$

So,

$= > \frac{1}{2} \times 200 \times {x}^{2} = 1 \\ \\ = > 100 \times {x}^{2} =1 \\ \\ {x}^{2} = \frac{1}{100} \\ \\ = >x= \frac{1}{10} =0.1$

Hence the total compression in the string will be 0.1 m

{hope it helps}