Question

Two identical steel cubes each of mass 50 g and side 1 cm collide head on face to face with a speed of $10 \mathrm {~ms} ^ { - 1 }$. The maximum compression of each cube is:

($Y _ { \text {steel } } = 2 \times 10 ^ { 11 } \mathrm { Nm } ^ { 2 }$)

• A. $5 \times 10 ^ { - 5 } \mathrm {~m}$
• B. $5 \times 10 ^ { - 6 } \mathrm {~m}$
• C.
• D.

Answer 1

Answer: (C)

Answer 2

Here, m = 50 g = $50 \times 10 ^ { - 3 } \mathrm {~kg}$

$\mathrm { Y } = 2 \times 10 ^ { 11 } \mathrm { Nm } ^ { - 2 }$

According to Hooke’s law

or $F = \frac { Y A \Delta L } { L }$ ……(i)

Where A is the surface area and L is the length of the cube.

If k is the compression of the cube, then

$\mathrm { F } = \mathrm { k } \Delta \mathrm { L }$

(Using (i))

$\Rightarrow \mathrm { K } = \frac { \mathrm { YA } } { \mathrm { L } } = \mathrm { YL }$ …(ii)

Initial kinetic energy

= $2 \times \frac { 1 } { 2 } m v ^ { 2 } = m v ^ { 2 }$

Final potential energy

$= 2 \times \frac { 1 } { 2 } \mathrm { k } ( \Delta \mathrm { L } ) ^ { 2 } = \mathrm { k } ( \Delta \mathrm { L } ) ^ { 2 }$

According to the mechanical energy conservation

(Using (ii))

$= \sqrt { \frac { 5 } { 2 } \times 10 ^ { - 13 } } = \sqrt { 2.5 \times 10 ^ { - 13 } } = 5 \times 10 ^ { - 7 } \mathrm {~m}$