Chapter 7, Problem 63P

(a)

To determine

The change in momentum of $m_1$ during collision along x and y directions.

Answer to Problem 63P

The change in momentum of $m_1$ during collision along x and y directions are $-m_1{v}_i$ and $0.751m_1{v}_i$ respectively.

Explanation of Solution

Using the conservation of linear momentum along x direction,

    $\begin{array}{c}\Delta p_{1,x}=m_2\left(v_{2,x,i}-v_{2,x,f}\right)\\ =m_2\left(v_{2,x,i}-\frac{1}{4}{v}_i\cos \theta \right)\end{array}$                                                    

Using the conservation of linear momentum along y direction,

    $\begin{array}{c}\Delta p_{1,y}=m_2\left(v_{2,y,i}-v_{2,y,f}\right)\\ =m_2\left(v_{2,y,i}-\frac{1}{4}{v}_i\sin \theta \right)\end{array}$                                                   

Conclusion:

Substitute 0 m/s for $v_{2,x,i}$, $5m_1$ for $m_2$, $-36.9°$ for $\theta$ to get $\Delta p_{1,x}$.

    $\begin{array}{c}\Delta p_{1,x}=\left(5m_1\right)\left(0-\frac{1}{4}{v}_i\cos \left(-36.9°\right)\right)\\ =-0.9997m_1{v}_i\\ \approx -m_1{v}_i\end{array}$

Substitute 0 m/s for $v_{2,y,i}$, $5m_1$ for $m_2$, $-36.9°$ for $\theta$ to get $\Delta p_{1,y}$.

    $\begin{array}{c}\Delta p_{1,x}=\left(5m_1\right)\left(0-\frac{1}{4}{v}_i\sin \left(-36.9°\right)\right)\\ =0.751m_1{v}_i\end{array}$

Therefore, change in momentum of $m_1$ during collision along x and y directions are $-m_1{v}_i$ and $0.751m_1{v}_i$ respectively.

(b)

To determine

The change in momentum of $m_2$ during collision along x and y directions.

Answer to Problem 63P

The change in momentum of $m_2$ during collision along x and y directions are $m_1{v}_i$ and $-0.751m_1{v}_i$ respectively.

Explanation of Solution

Using the conservation of linear momentum along x direction,

    $\begin{array}{c}\Delta p_{2,x}=m_1\left(v_{1,x,i}-v_{1,x,f}\right)\\ =m_1\left(v_i-v_{1,x,f}\right)\end{array}$                                                   

Using the conservation of linear momentum along y direction,

    $\begin{array}{c}\Delta p_{2,y}=m_1\left(v_{1,y,i}-v_{1,y,f}\right)\\ =m_1\left(v_{1,y,i}-v_1\right)\end{array}$                                                   

Conclusion:

Substitute 0 m/s for $v_{1,x,f}$ to get $\Delta p_{2,x}$.

    $\begin{array}{c}\Delta p_{2,x}=m_1\left(v_i-0\right)\\ =m_1{v}_i\end{array}$

Substitute 0 m/s for $v_{1,y,i}$ and $0.751v_i$ for $v_1$ to get $\Delta p_{2,y}$.

    $\begin{array}{c}\Delta p_{2,y}=m_1\left(0-0.751v_i\right)\\ =-0.751m_1{v}_i\end{array}$

Therefore, change in momentum of $m_2$ during collision along x and y directions are $m_1{v}_i$ and $-0.751m_1{v}_i$ respectively. The momentum changes in (a) and (b) are equal in magnitude and opposite in direction.