Question

Block 'A' is hanging from a vertical spring and is at rest. Block 'B' strikes the block 'A' with velocity 'v' and sticks to it. Then the value of 'v' for which the spring just attains natural length is –

• A. $\sqrt { \frac { 60 \mathrm { mg } ^ { 2 } } { \mathrm { k } } }$
• B. $\sqrt { \frac { 6 \mathrm { mg } ^ { 2 } } { \mathrm { k } } }$
• C. $\sqrt { \frac { 10 \mathrm { mg } ^ { 2 } } { \mathrm { k } } }$
• D. None of these

Answer 1

Answer: (B)

$\sqrt { \frac { 6 \mathrm { mg } ^ { 2 } } { \mathrm { k } } }$

Answer 2

The initial extention in spring is X₀ = $\frac { \mathrm { mg } } { \mathrm { k } }$ .

Just after collision of B with A the speed of combined mass is $\frac { v } { 2 }$ .

For the spring to just attain natural length the combined mass must rise up by X₀ = $\frac { \mathrm { mg } } { \mathrm { k } }$ and comes to rest.

Applying conservation of energy between initial and final states

$\frac { 1 } { 2 }$ 2 m $\left[ \frac { \mathrm { v } } { 2 } \right] ^ { 2 }$ + $\frac { 1 } { 2 }$ k $\left[ \frac { \mathrm { mg } } { \mathrm { k } } \right] ^ { 2 }$ = 2mg $\left[ \frac { \mathrm { mg } } { \mathrm { k } } \right]$

solving we get v = $\sqrt { \frac { 6 \mathrm { mg } ^ { 2 } } { \mathrm { k } } }$ .