# 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.

P Prateek Shrivastava

As we learnt in

Perfectly Inelastic Collision -

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

- wherein

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

$\fn_jvn m_{1},m_{2}= masses$

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

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

$(100 kg + 90kg )V=20 kg$

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

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