Chapter 7, Problem 71GP

To determine

The speeds of billiard balls after collision

Answer to Problem 71GP

  $121\frac{\text{m}}{\text{s}}$

Explanation of Solution

Given:

Mass of the block $=M=1.35\text{kg}$

Mass of the bullet $=m=25\text{g}=0.025\text{kg}$

Initial velocity of bullet before collision $=v_i$

Initial velocity of block before collision $=V_i=0\frac{\text{m}}{\text{s}}$

Initial velocity of bullet block combination after collision $=V_{si}$

Final velocity of bullet block combination after it stops $=V_{sf}=0\frac{\text{m}}{\text{s}}$

Stopping distance for bullet block combination after collision $=d=9.5\text{m}$

Angle between frictional force and distance moved $=\theta =180°$

Formula Used:

Normal force is given as

  $\begin{array}{l}N=mg\\ \text{Where }N\text{ = normal force, }m\text{ = mass of the object}\end{array}$

Force due to frictional force is given as

  $f=\mu N$

Work done by a force is given as

  $W=Fd\cos \theta$

According work-change in kinetic energy theorem,

  $\begin{array}{l}W=(0.5)m(v_f^2-v_i^2)\\ \text{Where }m\text{ = mass , }{v}_i=\text{initial speed, }{v}_f=\text{final speed}\end{array}$

According conservation of momentum principle, for perfectly elastic collision,

  $m_1{v}_{1i}+m_2{v}_{2i}=m_1{v}_{1f}+m_2{v}_{2f}$

Calculation:

Normal force acting on the combination of bullet and block after collision is given as

  $N=(m+M)g$

Frictional force acting on the bullet-block combination after collision is given as

  $\begin{array}{l}f=\mu N\\ f=\mu (m+M)\end{array}$

Using work-change in kinetic energy theorem for the motion of bullet-block combination after the collision,

  $\begin{array}{l}W=(0.5)(m+M)(V_{sf}^2-V_{si}^2)\\ fd\cos \theta =(0.5)(m+M)(V_{sf}^2-V_{si}^2)\\ \mu (m+M)d\cos \theta =(0.5)(m+M)(V_{sf}^2-V_{si}^2)\\ \mu d\cos \theta =(0.5)(V_{sf}^2-V_{si}^2)\\ (0.25)(9.5)\cos 180=(0.5)(0^2-V_{si}^2)\\ V_{si}=2.2\frac{\text{m}}{\text{s}}\end{array}$

Using conservation of momentum for the collision of bullet and block

  $\begin{array}{l}m{v}_i+M{V}_i=(m+M)V_{si}\\ (0.025)v_i+(1.35)(0)=(0.025+1.35)(2.2)\\ v_i=121\frac{\text{m}}{\text{s}}\end{array}$

Conclusion:

Therefore, muzzle speed of the bullet is $121\frac{\text{m}}{\text{s}}$