Question

Two bodies of masses $0.1kg$ and $0.4kg$ move towards each other with velocities $1m/s$ and $0.1m/s$ respectively. After collision they stick together. In $10s$ the combined mass travels

• A. $120m$
  
• B. $0.12m$
  
• C. $12m$
  
• D. $1.2m$
  

Answer 1

The correct option is D $1.2m$

$\text{Step 1: Initial \& final situation [Ref. Fig.]}$

$\text{Step 2: Apply momentum conservation}$

Since net external force on the system is zero, therefore applying conservation of momemtum

Considering rightward direction as positive

$P_i=P_f$

$\Rightarrow m_1{u}_1-m_2{u}_2=(m_1+m_2)V$

$\Rightarrow V=\left(\frac{m_1{u}_1-m_2{u}_2}{m_1+m_2}\right)$

$\text{Step 3: Calculate distance traveled}$,

As the acceleration is zero, therefore the distance traveled will be

$d=Vt=\left(\frac{m_1{u}_1-m_2{u}_2}{m_1+m_2}\right)t$

Putting given values:-

$d=\left(\frac{0.1\left(kg\right)\times 1\left(m/s\right)-0.4\left(kg\right)\times 0.1\left(m/s\right)}{0.1\left(kg\right)+0.4\left(kg\right)}\right)\times 10\left(s\right)$

$\Rightarrow d=1.2m$

Hence option $D$ is correct.