A 90 kg person is sitting in a boat at rest with a mass of 100 kg. In the boat is a stone with a mass of 5.0 kg. The person throws the stone at 4.00 m/s horizontally in the NORTH direction. The velocity of the person and the boat after the throwing the stone is
 

  • Option 1)

    2.00 m/s NORTH

  • Option 2)

    1.21 m/s SOUTH.

  • Option 3)

    1.21 m/s NORTH.

  • Option 4)

    0.11 m/s SOUTH.

 

Answers (1)
P Prateek Shrivastava

As we learnt in

Perfectly Inelastic Collision -

v= \frac{m_{1}v_{1}+m_{2}v_{2}}{m_{1}+m_{2}}

- wherein

Two bodies stick together after the collision ,so there will be a final common velocity (v)

m_{1},m_{2}= masses

v_{1}=\: initial \: velocity \: mass\: m_{1}

v_{2}=\: initial \: velocity \: mass\: m_{2}

 

 (100 kg + 90kg )V=20 kg

where 20 kg is Momentum

190 V=20=V=\frac{20}{190}=0.11 ms^{-1}

 

 


Option 1)

2.00 m/s NORTH

This option is incorrect

Option 2)

1.21 m/s SOUTH.

This option is incorrect

Option 3)

1.21 m/s NORTH.

This option is incorrect

Option 4)

0.11 m/s SOUTH.

This option is correct

Exams
Articles
Questions