Chapter 8, Problem 1OQ

To determine

The true statement from the given options.

Answer to Problem 1OQ

Option (d).

Explanation of Solution

Write the expression for momentum,

   $P=mv$        (I)

Here, $P$ is the momentum, $m$ is the mass, and $v$ is the velocity.                                                

Write the expression for the conservation of momentum that is the momentum is the same before and after the collision,

  $m_{1i}{v}_{1i}\text{}+m_{2i\text{}}{v}_{2i}+\dots =m_{1f}{v}_{1f}\text{}+m_{2f\text{}}{v}_{2f}+\dots$        (I)

Here, here subscripts $i$  and $f$  denote the initial and final momenta of objects in a system, $m_1$ and $v_1$ are the mass and velocity respectively for object $1$ and $m_2$ , $v_2$ are the mass and velocity respectively for object $2$.

Momentum is the same before the collision, thus expression (I) can be re-written as,

  $m_1{v}_1+m_2{v}_2$        (II)

Here, $m_1$ and $v_1$ are the mass and velocity respectively for object $1$ and $m_2$ , $v_2$ are the mass and velocity respectively for object $2$.

Conclusion

Substitute $\text{3 kg}$ for $m_1$ and $\text{+2 m/s}$ for $v_1$ and $\text{2 kg}$ for $m_2$ and $\text{-4 m/s}$ for $v_2$ in expression (II).

  $\begin{array}{c}{m}_1{v}_1+m_2{v}_2=\left(\text{3 kg}\right)\left(\text{+2 m/s}\right)+\left(\text{2 kg}\right)\left(\text{-}\text{4 m/s}\right)\\ =6-8\\ =-2\text{kg}\text{.m/s}\text{.}\end{array}$

Since, momentum is the same before and after the collision, the momentum of the system will be $-2\text{kg}\text{.m/s}\text{.}$ thus, option (d) is the correct option.

All other option are incorrect.