Chapter 9, Problem 72A

To determine

The velocity of the launcher after the launch.

Answer to Problem 72A

  $4.9\text{m/s}$ in backward direction

Explanation of Solution

Given:

The mass of the projectile, $m_p=50.0\text{g = 0}\text{.05}\text{kg}$

The mass of the launcher, $m_l=4.65\text{kg}$

Initial velocity of launcher, $v_{li}=2.00\text{m/s}$

Final velocity of the projectile, $v_{pf}=647\text{m/s}$

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 both the launcher and projectile as a system, then it will be a closed and isolated system and law of conservation of momentum can be applied to it.

Also, the initial velocity of the projectile will be same as that of the launcher, thus

  $v_{pi}=2\text{m/s}$

Let the final speed of the launcher $v_{lf}$ , then using the law of conservation of linear momentum,

  $\begin{array}{c}{p}_i=p_f\\ m_l{v}_{li}+m_p{v}_{pi}=m_l{v}_{lf}+m_p{v}_{pf}\end{array}$

Using given values in above,

  $\begin{array}{c}\left(4.65\right)\left(2\right)+\left(0.05\right)\left(2\right)=\left(4.65\right)\left(v_{lf}\right)+\left(0.05\right)\left(647\right)\\ v_{lf}=\frac{\left(9.3+0.1-32.35\right)}{4.65}\text{m/s}\\ v_{lf}=-4.9\text{m/s}\end{array}$

Negative sign of the above velocity denotes that after the launch, the launcher moves in backward direction.

Conclusion:

Thus, the velocity of the launcher after the launch is $4.9\text{m/s}$ in backward direction.