Question

A light particle moving horizontally with a speed of $12m/s$ strikes a very heavy block moving in the same direction at $10m/s$. The collision is one-dimensional and elastic. If the speed of the block does not change after the collision, the particle will

• A. Move at $2m/s$ in its original direction
• B. Move at $8m/s$ in its original direction
• C. Move at $8m/s$ opposite to its original direction
• D. Move at $12m/s$ opposite to its original direction

Answer 1

The correct option is B Move at $8m/s$ in its original direction

Conservation of momentum,

$m_1{v}_1+m_2{v}_2=m_1{v}_1^{\prime }+m_2{v}_2^{\prime }$

$\Rightarrow m_1(v_1-v_1^{\prime })=m_2(v_2-v_2^{\prime })$.......(1)

Conservation of kinetic energy requires,

$\frac{1}{2}(m_1{v}_1^2+m_2{v}_2^2)=\frac{1}{2}(m_1{v_1^{\prime }}^2+m_2{v_2^{\prime }}^2)$

$\Rightarrow m_1(v_1^2-{v_1^{\prime }}^2)=m_2(v_2^2-{v_2^{\prime }}^2)$

$\Rightarrow m_1(v_1-v_1^{\prime })(v_1+v_1^{\prime })=m_2(v_2-v_2^{\prime })(v_2+v_2^{\prime })$.........(2)

dividing (2) and (1)

$v_1+v_1^{\prime }=v_2+v_2^{\prime }$

$12+v_1^{\prime }=10+10$

$v_1^{\prime }=8m/s$