Answer to Question #149064 in Physics for mahira

Question

A toy bird of mass 2kg is
being hung from the ceiling with the help of a 0.1m rope. A bullet of mass 0.01kg hits the bird horizontally and
sticks to it. If the bird afterwards, at its highest vertical displacement touches the ceiling anyway, what was the
speed of the bullet?

Answer 1

Since the bird (with bullet inside) with mass $m_1 = 2kg+0.01kg = 2.01kg$ rised to the height $h = 0.1m$ after the hit, its gain in potential energy is:

$$U = m_1gh = 2\cdot 9.81\cdot 0.1 = 1.926\space J$$

According to the energy conservation law, the initial kinetic energy (just after impact) of the bird was the same:

$$K = U = 1.926\space J$$

On the other hand, the kinetic energy is equal to:

$$K = \dfrac{p_1^2}{2m_1}$$

where $p_1$ is the initial momentum of the bird. Expressing $p_1,$ obtain:

$$p_1 = \sqrt{2m_1K}$$

Finally, according to the momentum conservation, the initial momentum of the bird should be equal to the momentum of the bullet:

$$p_1 = m_2v$$

where $m_2 = 0.01kg$ is the mass and $v$ is the speed of the bullet. Combining all it together:

$$m_2v = \sqrt{2m_1K}\\ v=\dfrac{\sqrt{2m_1K}}{m_2} = \dfrac{\sqrt{2\cdot 2.01\cdot 1.926}}{0.01} \approx 278m/s$$

Answer. 278 m/s.