Question

A car of mass 1050 kg is moving towards east with a speed of 15 $m{s}^{-1}$. A truck is moving towards west. The mass of the truck is 6320 kg. both the vehicle collide head-on and remain locked together after the collision. Based on above information, answer the following question.

If the truck is moving at 10 $m{s}^{-1}$ before collision then the common velocity of the two vehicles after collision is

• A. 6.44 $m{s}^{-1}$ towards west
• B. 10.7 $m{s}^{-1}$ towards east
• C. 6.44 $m{s}^{-1}$ towards east
• D. 10.7 $m{s}^{-1}$ towards west

Answer 1

The correct option is A

6.44 $m{s}^{-1}$ towards west

Given:

Mass of car $m_1$=1050 kg

Mass of truck $m_2$=6320 kg

Taking the west direction to be positive and the east direction negative

velocity of car $v_1$= $-15\hat{i}m{s}^{-1}$

After collision they remain locked so, we have to consider inelastic collision

Applying Conservation of momentum

If the truck has velocity $v_2$=$-10\hat{i}m{s}^{-1}$ before collision then the common velocity of the two vehicles after the collision is calculated using the concept of momentum

$$\begin{gathered} -m_1{v}_1+m_2{v}_2=(m_1+m_2)v \\ v=\frac{-m_1{v}_1+m_2{v}_2}{m_1+m_2}=\frac{(-1050\times 15\hat{i})+(6320\times 10\hat{i})}{1050+6320} \end{gathered}$$

$V=6.44\hat{i}m{s}^{-1}$

Therefore the common velocity of the two vehicles after the collision is 6.44 m/s toward west

Hence option (a) is the right answer