# What is the momentum of a system of two particles with these respective masses and velocities: 3.76 kg moving north at 5.6 m/s and 4.2 kg moving northwest at 2.3 m/s? The momentum of this system will be the vector sum of the momenta of the two particles:

`vecp = m_1vecv_1 + m_2 vecv_2`

To find a vector, we need to find its magnitude and direction, or its components along the coordinate axis. We can choose the coordinate system such that...

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The momentum of this system will be the vector sum of the momenta of the two particles:

`vecp = m_1vecv_1 + m_2 vecv_2`

To find a vector, we need to find its magnitude and direction, or its components along the coordinate axis. We can choose the coordinate system such that the vertical (y) axis is pointing North and the horizontal (x) axis pointing East.

Since the first particle is moving North, its momentum has no horizontal component. The second particle is moving Northwest, which means its momentum makes an equal angle (45 degrees) with the positive y-axis and negative x-axis. So, its x-component is

`m_2v_(2x) =-m_2v_2*cos(45) = -m_2v_2*sqrt(2)/2`

Then the x-component of total momentum is

`p_x = 0- m_2v_2*sqrt(2)/2 = -4.2*2.3*sqrt(2)/2 = -6.83 kg*m/s`

The y-component of the momentum of the first particle equals its magnitude:

` m_1v_(1y) = m_1v_1 = 3.76*5.6 = 21.06 kg*m/s`  and the y-component of the momentum of the second particle is

`m_2v_(2y) = m_2v_2cos(45) = 4.2*2.3*sqrt(2)/2 = 6.83 kg*m/s`

Then the y-component of the total momentum is

`p_y = m_1v_(1y) + m_2v_(2y) = 27.9 kg*m/s`

The momentum of the given system, in component form, is

(-6.83 kg*m/s, 27.9 kg*m/s)

The x-component is negative and y-component is positive, so the momentum is pointing at some angle North of West.

To find its magnitude, we can use Pythagorean theorem:

`p = sqrt(p_x^2 + p_y^2) = 28.7 kg*m/s`

The angle between the momentum and negative direction of x-axis is given by

`theta = tan^(-1) |p_y|/|p_x|= 76.2` degrees.

The momentum of the given system has the magnitude of 28.7 kg*m/s and is directed 76.2 degrees North of West.

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