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Two metal balls of masses $100 g$ and $200 g$ are moving along the same line and direction with velocities of $2 m s^{- 1}$ and $1 m s^{- 1}$ , respectively. The two balls collide and after the collision, the first ball moves at a velocity of $1.67 m s^{- 1}$ . What will be the velocity of the second ball?

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1.165 m s^{- 1}

1.155 m s^{- 1}

1.265 m s^{- 1}

1.123 m s^{- 1}

answer is A.

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Detailed Solution

The velocity of the second ball will be

1.165 m s^{- 1}

.
To find the solution we will have to calculate the momentum of the first ball and the second ball separately.
Momentum of the first ball:
Momentum of first ball

=

mass

\times

velocity

\Rightarrow

= \frac{100}{1000} kg \times 2 m s^{- 1}

Momentum of first ball

= 0.2 kg \frac{m}{s}

Now to find the Momentum of the second ball
Momentum of the second ball

=

mass

\times

velocity

\Rightarrow = \frac{200}{1000} kg \times 1 m s^{- 1}

Momentum of the second ball

= 0.2 kg m s^{- 1}

Total momentum

= 0.2 + 0.2

Total momentum

= 0.4 kg m s - 1

After collision, the velocity of first ball of mass

100 g

becomes

1.67 m s^{- 1}

.
So, momentum of first ball (after collision)

= \frac{100}{1000} kg \times 1.67 m s^{- 1}

Momentum of first ball

= 0.167 kg m s^{- 1}

After collision, the velocity of Second ball of mass

200 g

becomes

v m s^{- 1}

.
So, momentum of Second ball (after collision)

= \frac{200}{1000} kg \times v m s^{- 1}

momentum of Second ball

= 0.2 v kg m s - 1

Total momentum (after collision)

= 0.167 + 0.2 v

Now, according to the law of conservation of momentum:
Total momentum before collision

=

Total momentum after collision

\Rightarrow 0.4 = 0.167 + 0.2 v

\Rightarrow 0.2 v = 0.4 - 0.167

\Rightarrow v = \frac{0.233}{0.2}

\Rightarrow v = 1.165 m s^{- 1}

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