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Q46PE

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Found in: Page 291

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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

In both the cases momentum is conserved.

See the step by step solution

## Step 1: Definition of Linear Momentum

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$${\theta _2} = 45^\circ$$.

## Step 2: Linear momentum along X-axis

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

## Step 3: Linear momentum along Y-axis

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.

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