Chapter 9, Problem 71A

To determine

The velocity of the $2.0\text{kg}$ cart after the two lab carts released.

Answer to Problem 71A

  $0.3\text{m/s}$ in the opposite direction of 1^st cart’s final motion

Explanation of Solution

Given:

The mass of the 1^st cart, $m_1=5.0\text{kg}$

The mass of the 2^nd cart, $m_2=2.0\text{kg}$

Final velocity of 1^st cart after the release, $v_{1f}=0.12\text{m/s}$

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 carts 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 final direction of 1^st cart’s motion as positive. And, since initially both the carts was at rest, hence,

  $v_{1i}=v_{2i}=0\text{m/s}$

Let, the final velocity of the $2.0\text{kg}$ cart, that is 2^nd cart be $v_{2f}$ , then using the law of conservation of linear momentum,

  $\begin{array}{c}{p}_i=p_f\\ m_1{v}_{1i}+m_2{v}_{2i}=m_1{v}_{1f}+m_2{v}_{2f}\end{array}$

Using given values in above,

  $\begin{array}{c}\left(5\right)\left(0\right)+\left(2\right)\left(0\right)=\left(5\right)\left(0.12\right)+\left(2\right)\left(v_{2f}\right)\\ v_{2f}=-0.3\text{m/s}\end{array}$

Negative sign of the above velocity denotes that after the collision, the second cart moves in the opposite direction of 1^st cart’s final motion.

Conclusion:

Thus, the final velocity of the $2.0\text{kg}$ cart is $0.3\text{m/s}$ in the opposite direction of 1^st cart’s final motion.