Chapter 8, Problem 58A

A 1-kg dart moving horizontally at 10 m/s strikes and sticks to a wood block of mass 9 kg, which slides across a friction-free level surface. What is the speed of the block and the dart after the collision?

To determine

To calculate: the speed of the block and the dart.

Answer to Problem 58A

  $1\text{m/s}$

Explanation of Solution

Given info:

The mass of the dart is 1kg and its velocity before collision is $10\text{m/s}$ .

The mass of the wooden block is 9kg.

Formula used:

According to the law of conservation of momentum,

Initial momentum = final momentum

Momentum:

P=mv

As per law of conservation of momentum is:

  $m_1{v}_{1b}+m_2{v}_{2b}=m_1{v}_{1a}+m_2{v}_{2a}$

Where, $m_1\text{and}{m}_2$ are masses of blob and $v_{1b}\text{and}{v}_{2b}$ are velocities of the particles before collision and $v_{1a}\text{and}{v}_{2a}$ are the velocities after collision.

Calculation:

According to the law of conservation of momentum,the system momentum remains conserved if no external force is acting on the system. After the collision, the velocities become equal. So, the formula becomes,

  $m_1{v}_{1b}+m_2{v}_{2b}=\left(m_1+m_2\right)v_A$

By substituting the values in the formula, the velocity after the collision is

  $\begin{array}{l}1\text{kg×}10\text{m/s}+0=\left(1+9\right)v_A\\ \Rightarrow v_A=\frac{10}{10}=1\text{m/s}\end{array}$

Conclusion:

Thus, the speed of the block and dart after collision is $1\text{m/s}$ .