Question

Two bodies of mass $0.4\text{kg}$ and $0.8\text{kg}$ move towards each other with initial velocities $5\text{m/s}$ and $2\text{m/s}$ respectively. After the collision, they stick together. Then, find the distance travelled by the combined mass, $12\text{sec}$ after the collision.

• A. $3\text{m}$
• B. $12\text{m}$
• C. $8\text{m}$
• D. $4\text{m}$

Answer 1

The correct option is D $4\text{m}$

Let $m_1=0.4\text{kg},m_2=0.8\text{kg}$
The initial velocities of bodies before collision are $v_1=+5\text{m/s},v_2=-2\text{m/s}$ as shown in figure.


After the collision, masses stick together, so combined mass is $(m_1+m_2)$. Let the masses move with common velocity $v$.


Applying conservation of linear momentum on system of bodies:
$P_i=P_f$
$m_1{v}_1+m_2{v}_2=(m_1+m_2)v$
$\Rightarrow \left(0.4\right)(+5)+\left(0.8\right)(-2)=\left(1.2\right)v$
$\therefore v=\frac{0.4}{1.2}=\frac{1}{3}\text{m/s}$

Hence, the combined system moves with velocity $v$ in the initial direction of motion of $0.4\text{kg}$ mass.
$\Rightarrow$Distance travelled by combined mass in time interval $t=12\text{sec}$, after collision is
$d=v\times t=\frac{1}{3}\times 12=4\text{m}$