Confirm that the results of the example 8.7and do conserve momentum in both the x- and y-directions

The product of an object's mass and velocity is called linear momentum, translational momentum, or simply momentum. It's a two-dimensional vector quantity with a magnitude and a direction.

The mass of the first one is\({m_1} = 0.250\;kg\).

The mass of the second one is\({m_2} = 0.400\;kg\).

The initial velocity of the first one before collision is\({\theta _2} = 45^\circ \).

We can write along X-axis,

Linear momentum before collision is,

\(\begin{array}{c}mv = 0.250 \times 2\\ = 0.500\;{\rm{kg}} \cdot {\rm{m/s}}\end{array}\)

Linear momentum after collision is,

\(\begin{array}{l}{m_1}{{v'}_1}\cos {\theta _1} + {m_2}{{v'}_2}\cos {\theta _2}\\ = \left[ {0.250 \times 1.50 \times \cos 55^\circ + 0.400 \times 896 \times \cos 312^\circ } \right]\;{\rm{kg}} \cdot {\rm{m/s}}\\ = 0.500\;{\rm{kg}} \cdot {\rm{m/s}}\end{array}\)

Both are the same

We can write along Y-axis,

Linear momentum before collision is zero as the particles are moving along the X-axis.

Linear momentum after collision is,

\(\begin{array}{l}{m_1}{{v'}_1}\sin {\theta _1} + {m_2}{{v'}_2}\sin {\theta _2}\\ = \left[ {0.250 \times 1.50 \times \sin 55^\circ + 0.400 \times 896 \times \sin 312^\circ } \right]\;{\rm{kg}} \cdot {\rm{m/s}}\\ = 0\end{array}\)

Both are the same.