Two blocks of masses 3kg and 6kg respectively are placed on a smooth horizontal surface. They are connected by a light spring of force constant k = 200 N/m. Initially the spring is unstretched. The indicated velocities are imparted to the blocks. The maximum extension of the spring will be

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

20 cm

25 cm

15 cm

30 cm

answer is A.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

From conservation of momentum,

Initial momentum = momentum of center of mass

$m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) V_{cm} \Rightarrow 6 \times 2 - 3 \times 1 = (6 + 3) V_{cm} \Rightarrow V_{cm} = 1 m / s$

$From conservation of energy,$

$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} {kx}^{2} + \frac{1}{2} (m_{1} + m_{2}) V_{cm}^{2}$

$\begin{array}{l} \Rightarrow \frac{1}{2} (3) (1)^{2} + \frac{1}{2} (6) (2)^{2} = \frac{1}{2} {Kx}^{2} + \frac{1}{2} (9) V^{2} \end{array}$

$\Rightarrow \frac{3}{2} + \frac{24}{2} = \frac{1}{2} (200) x^{2} + \frac{1}{2} 9 (1)^{2}$

$∴ 100 x^{2} = 9 \Rightarrow x = \sqrt{\frac{9}{100}} = 0 .3 m$

Watch 3-min video & get full concept clarity

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test