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# Help me understand! A 4.00 kg ball is moving at 2.00 m/s to the WEST and a 6.00 kg ball is moving at 2.00 m/s to the NORTH. The total momentum of the system is

A 4.00 kg ball is moving at 2.00 m/s to the WEST and a 6.00 kg ball is moving at 2.00 m/s to the NORTH. The total momentum of the system is

• Option 1)

21.6 kg m/s at an angle of 17.7 degrees NORTH of WEST.

• Option 2)

14.4 kg m/s at an angle of 45.2 degrees SOUTH of WEST.

• Option 3)

21.6 kg m/s at an angle of 45.2 degrees SOUTH of WEST.

• Option 4)

14.4 kg m/s at an angle of 56.3 degrees NORTH of WEST.

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As discussed in

Perfectly Elastic Collision -

Law of conservation of momentum and that of Kinetic Energy hold good.

- wherein

$\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}= \frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}$

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$m_{1},m_{2}:masses$

$u_{1},v_{1}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{1}$

$u_{2},v_{2}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{2}$

$P=-4\times 2\hat{i} + 6\times 2\hat{J} = -8\hat{i}+12\hat{J}$

Magnitude of $P = \sqrt{8^{2}+12^{2}} = \sqrt{64+144} = 14.4 \:\: kg \:\: ms^{-1}$

$\therefore tan\theta = \frac{12}{-8} \Rightarrow\theta = \tan ^{-1} \left ( \frac{12}{8} \right ) = -56.3+\pi$

$= 56.3^{\circ} N$ of  west

Option 1)

21.6 kg m/s at an angle of 17.7 degrees NORTH of WEST.

This solution is incorrect.

Option 2)

14.4 kg m/s at an angle of 45.2 degrees SOUTH of WEST.

This solution is incorrect.

Option 3)

21.6 kg m/s at an angle of 45.2 degrees SOUTH of WEST.

This solution is incorrect.

Option 4)

14.4 kg m/s at an angle of 56.3 degrees NORTH of WEST.

This solution is correct.

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