Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 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.
     

 

Answers (1)

best_answer

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.

Posted by

divya.saini

View full answer

Similar Questions

Trending Articles/News

anam-khan

BITSAT 2024 application correction facility begins tomorrow; editable fields
anam-khan

BITSAT 2024 registration window closes tomorrow; exam from May 20
anam-khan

BITSAT 2024 registration deadline extended till April 16; session 1 exam from May 19 to 24

Latest Question