Chapter 9.2, Problem 20PP

To determine

The speed bag when the bullet exits.

Answer to Problem 20PP

  $2.8\text{m/s}$

Explanation of Solution

Given:

The mass of the bullet, $m_b=35.0\text{g}=0.035\text{kg}$

The mass of the stationary bag of flour, $m_B=2.5\text{kg}$

The initial speed of the bullet is, $v_{bi}=475\text{m/s}$

The bullet passes through the bag and exits at a speed of, $v_{bf}=275\text{}\text{m/s}$

Formula used:

Momentum of an object of mass m and moving with velocity v is given by,

  $p=mv$

And, law of conservation of linear momentum: Momentum is conserved in a closed and isolated system and is given by,

  $p_i=p_f$

Calculation:

Consider the bullet and bag as a system, then it will be a closed and isolated system and law of conservation of momentum can be applied to it.

Now, since the bag of floor was at rest initially, thus initial velocity of bag is,

  $v_{Bi}=0\text{m/s}$

Let, the speed of the bag after the bullet passes through it be $v_{Bf}$ , then using the law of conservation of linear momentum,

  $\begin{array}{c}{p}_i=p_f\\ m_B{v}_{Bi}+m_b{v}_{bi}=m_B{v}_{Bf}+m_b{v}_{bf}\end{array}$

Using given values in above,

  $\begin{array}{c}\left(2.5\right)\left(0\right)+\left(0.035\right)\left(475\right)=\left(2.5\right)v_{Bf}+\left(0.035\right)\left(275\right)\\ v_{Bf}=2.8\text{m/s}\end{array}$

Conclusion:

Thus, the speed of the bag after the bullet left it is $2.8\text{m/s}$ .