Chapter 9, Problem 2STP

To determine

To Choose: The correct option.

Answer to Problem 2STP

C. $1.6\text{m/s}$

Explanation of Solution

Given:

The mass of the ice-skater, $m_I=40.0\text{kg}$

The mass of the sled, $m_s=10.0\text{kg}$

Initial velocity of ice-skater, $v_{Ii}=2.0\text{m/s}$

Since the sled was at rest initially, thus the initial velocity of the sled, $v_{si}=0\text{m/s}$

After collision, ice-skater and sled moves together.

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 ice-skater and sled as a system, then it will be a closed and isolated system and law of conservation of momentum can be applied to it.

Assume the initial direction of motion of ice-skater as positive.

Let, the final speed of the ice-skater and sled be $v_f$ , then using the law of conservation of linear momentum,

  $\begin{array}{c}{p}_i=p_f\\ m_I{v}_{Ii}+m_s{v}_{si}=\left(m_I+m_s\right)v_f\end{array}$

Using given values in above,

  $\begin{array}{c}\left(40\right)\left(2\right)+\left(10\right)\left(0\right)=\left(40+10\right)v_f\\ v_f=1.6\text{m/s}\end{array}$

Conclusion:

Thus, the speed of the ice-skater and sled after they collided is $1.6\text{m/s}$ .