As shown in the figure, a smooth rod is mounted just above a table top. A 10 kg collar, which is able to slide on the rod with negligible friction is fastened to a spring whose other end is attatched to a pivot at O. The spring has negligible mass, a relaxed length of 10 cm and a spring constant of 500 N/m. The collar is released frm rest at point A. Here, in the figure, BC = 20 cm, AB = 40 cm, and B0 = 30 cm.

What is it's velocity as it passes point B and C.???

Here is the diagram

Dear Sanchit

Calculate the length AO =50 CM

spring natural lenghth is given 10 cm

so springh is stretched by 50-10 = 40 cm

Total energy at point A=(1/2)*500*(.40)²

                                =40 Nm

At point B Spring Stretched by =30-10 =20 cm

 Total energy at point B=(1/2)*500*(.20)²+(1/2)*10*v²

                                 =10 +5v²  

 

At point C Spring Stretched by =26 cm

Total energy at point C=(1/2)*500*(.26)²+(1/2)*10*v²

Now first equate energy at A and B and find the velocity at B

Similarly equate energy at A and c and find the velocity at c

---

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.
 We are all IITians and here to help you in your IIT JEE preparation.

 All the best Sanchit
 
Regards,
Askiitians Experts
Badiuddin