Question

During the head on collision of two masses $1kg$ and $2kg$ the maximum energy of deformation is $\frac{100}{3}J$. If before collision the masses are moving in the same direction, then their velocity of approach before the collision is:

• A. $20\text{m/sec}$
• B. $10\sqrt{2}\text{m/sec}$
• C. $10\text{m/sec}$
• D. $5\text{m/sec}$

Answer 1

The correct option is C $10\text{m/sec}$

Find the common velocity after collision.
Formula used:
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

Applying momentum conservation,

$m_1{u}_1+m_2{u}_2=m_{1}v+m_{2}v$

$v=\frac{m_1{u}_1+m_2{u}_2}{m_1+m_2}$

Find the deformation.
Applying energy conservation:

$=\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2$

$=\frac{1}{2}(m_1+m_2)(v{)}^2+\Delta U_{deformation}$

$\Delta U_{deformation}=\frac{1}{2}\frac{m_1{m}_2}{(m_1+m_2)}\times (u_1-u_2{)}^2$

$=\frac{100}{3}$

$\Rightarrow u_1-u_2=10\text{m/sec}$
Final answer: (a)