Question

Two blocks A and B of mass m and 2m are connected by a mass less spring of force constant k. They are placed on a smooth horizontal plane. Spring is stretched by an amount x and then released. The relative velocity of the blocks when the spring comes to its natural length is –

• A. $\left( \sqrt { \frac { 3 \mathrm { k } } { 2 \mathrm {~m} } } \right) \mathrm { x }$
• B. $\left( \sqrt { \frac { 2 \mathrm { k } } { 3 \mathrm {~m} } } \right) \mathrm { x }$
• C. $\sqrt { \frac { 2 \mathrm { kx } } { \mathrm { m } } }$
• D. $\sqrt { \frac { 3 \mathrm {~km} } { 2 \mathrm { x } } }$

Answer 1

Answer: (A)

$\left( \sqrt { \frac { 3 \mathrm { k } } { 2 \mathrm {~m} } } \right) \mathrm { x }$

Answer 2

From conservation of mechanical energy

$\frac { 1 } { 2 }$ kx² = $\frac { 1 } { 2 }$ $\mu \mathrm { v } _ { \mathrm { r } } ^ { 2 }$ ... (1)

Here, m = reduced mass of the blocks

= $\frac { ( m ) ( 2 m ) } { m + 2 m }$ = $\frac { 2 } { 3 } \mathrm {~m}$

and v_r = relative velocity of the two.

Substituting in Equation (1), we get

kx² = $\operatorname { mv } _ { \mathrm { r } } ^ { 2 }$

\ v_r = $\left( \sqrt { \frac { 3 \mathrm { k } } { 2 \mathrm {~m} } } \right) \mathrm { x }$