Chapter 2, Problem 2.1P

Two blocks of mass 0.1 kg and 0.2 kg approch each other on a frictionless surface at velocities of 0.4 and 1 m/s, respectly. If the blocks collide and remain together, calculate their joint after the collision.

To determine

The joint velocity after the collision Two blocks of mass 0.1 kg and 0.2 kg approach each other on a frictionless surface at velocities of 0.4 and 1 m/s, respectively, if the blocks collide and remain together?

Explanation of Solution

Given info:

  $\begin{array}{l}{m}_1=0.1kg\\ m_2=0.2kg\\ v_1=0.4m/s\\ v_2=1m/s\end{array}$

Formula used:

By law of conservation of momentum,

  $m_1{v}_1+m_2{v}_2=MV$

Calculation:

Here momentum is conserved

Assume mass of first block and second block as $m_1$ and $m_2$ , and their respective velocities as $v_1$ and $v_2$ .After collision let the velocity and mass be M and V.

We have $v_1$ and $v_2$ are in opposite direction, so

  $\begin{array}{l}{m}_1{v}_1-m_2{v}_2=MV\\ 0.1\times 0.4-0.2\times 1=0.3\times V\\ V=\frac{0.04-0.2}{0.3}=\frac{-0.16}{0.3}=-\text{0}\text{.53m/s}\end{array}$

Conclusion:

Thus, the velocity of blocks after collision is 0.53 m/s towards left.