Consider two blocks A and B of masses = 20 kg and = 5 kg that are placed on a frictionless table. The block A moves with a constant speed v = 3 m/s towards the block B kept at rest. A spring with spring constang k = 3000 N/m is attached with the block B

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Consider two blocks A and B of masses $m_{1}$ = 10 kg and $m_{2}$ = 5 kg that are placed on a frictionless table. The block A moves with a constant speed v = 3 m/s towards the block B kept at rest. A spring with spring constant k = 3000 N/m is attached with the block B as shown in the figure. After the collision, suppose that the blocks A and B, along with the spring in constant compression state, move together, then the compression in the spring is, (Neglect the mass of the spring) [2025]

1 0.2 m

2 0.4 m

3 0.1 m

4 0.3 m

Ans.
(3)

Conservation of Linear momentum,

$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v_{cm}$

$v_{cm} \Rightarrow \frac{10 \times 3}{10 + 5} \Rightarrow \frac{30}{15} = 2 m / s$

Conservation of mechanical energy,

$\frac{1}{2} k x^{2} = \frac{1}{2} (10) {(3)}^{2} - [\frac{1}{2} (15) {(2)}^{2}]$

$\Rightarrow 3000 x^{2} = 90 - 60 = 30$

$\Rightarrow x^{2} = \frac{30}{3000} = \frac{1}{100}$

$\Rightarrow x = \frac{1}{10} m$

Want Us to Email you About Special Offers & Updates?