First slide

Simple hormonic motion

Question

Two blocks A and B each of mass m are connected by a mass less spring of natural length I and spring constant k. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length as shown in fig. (7). A third identical block C, also of mass

m, moving on the floor with a speed v along the line joining A and 8, and collides with A. Then

Moderate

A

the kinetic energy of the A-B system, at maximum compression of the spring, is zero
B

the kinetic energy of A-B system, at maximum compression of the spring, is (m v² / 4)
C

the maximum compression of the spring is $V \sqrt{(m / 2 k)}$
D

2 and 3 both are correct

Solution

Let the velocity acquired by,4 and B be v'
then $mv = {mv}^{'} = {mv}^{'} or v^{'} = v / 2 And \frac{1}{2} {mv}^{2} = \frac{1}{2} {mv}^{' 2} + \frac{1}{2} {mv}^{' 2} + \frac{1}{2} {kx}^{2} where x = displacement of the block {mv}^{2} = m {(\frac{v}{2})}^{2} + m {(\frac{v}{2})}^{2} + {kx}^{2} or \frac{{mv}^{2}}{2} = {kx}^{2} or x = v {(\frac{m}{2 k})}^{1 / 2} At maximum compression, the kinetic energy of AB system will be \frac{1}{2} m {(v^{'})}^{2} + \frac{1}{2} m (v^{' 2}) = m {(v^{'})}^{2} = m {(\frac{v}{2})}^{2} = \frac{{mv}^{2}}{4}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App