Chapter 7, Problem 58QAP

To determine

The velocities of the blocks of ice after the collision

Answer to Problem 58QAP

The velocities of the ice blocks after the collision are 5 m/s and 9 m/s.

Explanation of Solution

Given info:

A 10 kg block of ice is sliding due east at 8 m/s when it collides elastically with a 6 kg block of ice that is sliding in the same direction at 4 m/s.

Formula used:

Using the formula, conservation of linear momentum

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$ (1)

Here, $m_1\text{and }{m}_2$ are the masses of first and second balls respectively and $u_1\text{and }{u}_2$ are the velocities of the both balls before collision and $v_1\text{and }{v}_2$ are the velocities after collision.

Also, for elastic collision

  $e=\frac{v_2-v_1}{u_1-u_2}$ (2)

Here, e is coefficient of restitution is 1 for elastic collision.

Calculation:

We have, $m_1=10.00\text{kg, }{m}_2=6.00\text{kg and }{u}_1=8.00\text{m/s and }{u}_2=4\text{m/s}$

Substituting the given values in equ.(1), we get

  $\begin{array}{l}10\text{kg}\text{×}8\text{m/s + 6}\text{kg × 4 m/s}=10\text{kg}\times v_1+6\text{kg}\times v_2\\ \Rightarrow 10v_1+6v_2=104(3)\end{array}$

Now from equation(2)

  $\begin{array}{l}1=\frac{v_2-v_1}{8-4}(e=1,forelasticcollision)\\ \Rightarrow v_2-v_1=4(4)\end{array}$

From equations (3) and (4), we get

  $\begin{array}{l}10v_1+6(v_1+4)=104\\ 16v_1=104-24\\ \Rightarrow v_1=\frac{80}{16}=5\text{m/s}\end{array}$

Substituting in equation (4), we get

  $\begin{array}{l}(v_2-5)=4\\ \Rightarrow v_2=5+4=9\text{m/s}\end{array}$

Conclusion:

Thus, the velocities of the ice blocks after the collision are 5 m/s and 9 m/s.