Question

A 2.0 kg bird lands on a 1.0 x 10^1 kg bit of tree bark sitting on a frictionless ice-covered pond. The bird’s initial horizontal speed is 6.0 m/s. What is the final speed of the bird and bark?

Answer 1

Here in this type of question we can use momentum conservation

It is because we can see there is no external force on the system

So we can use momentum conservation principle

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

here we know that

$m_1 = 2 kg$

$v_{1i} = 6 m/s$

$m_2 = 1 * 10^1 kg$

$v_{2i} = 0$

now after bird sits on it then final speed of the both will be same

$v_{2f} = v_{1f} = v m/s$

$2*6 + 1*10^1 * 0 = (2 + 1* 10^1) * v$

$12 = 12*v$

$v = 1 m/s$

so final speed will be 1 m/s