Question

Pls help me out it’s urgent

Answer 1

Answer: 1.16 m/s

Here,

$m_{1} = 0.1 \: kg \\ m_{2} = 0.2 \: kg$

and,

$u_{1} = 2\: m/s \\ u_{2} = 1\: m/s \\ v_{1} = 1.67\: m/s$

Since, No External force is being applied to the system, Therefore, the momentum of the system is conserved.

Applying Conservation of Momentum, We get

$P_{i} = P_{f}$

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

$= > (0.1 \times 2) + (0.2 \times 1) = (0.1 \times 1.67) + (0.2 \times v_{2})$

$= > 0.2 + 0.2 = 0.167 + 0.2v_{2} \\ = >v_{2} = (0.4 - 0.167) \div 0.2 = \frac{0.233}{0.2} m/s$

Hence after collision, the velocity of second object is 1.16 m/s.

Hope this helps.