Question

A. O62.497 JB. O 40 JC. O 29.044 JD.44.934 JE. O48.953 J3. After a bullet of mass 0.055 kg is fired into aballistic pendulum of mass 4 kg. the bullet is embeddedin the pendulum and the pendulum rose to a height of 0.01 mCalulate the speed of the pendulum just after the collision. (1 point)A. O 0.443 m/sB. O 0.125 m/sC. O 0.713 m/sD.0.794 m/sE.0.344 m/s4. After a bullet of mass 0.03 kg is fired into aballistic pendulum of mass / kg, the bullet is embeddedin the pendulum and the pendulum rose to a height of 0.4 m.Calulate the speed with which the bullet was fired into theballistic pendulum. (1 point)A57 580 m/

Answer 1

Question 3.

Given:

• Mass of bullet = 0.055 kg

,

• Mass of pendulum = 4 kg

,

• Height = 0.01 m

Let's find the speed of the pendulum after the collision.

To find the speed of the collision, apply the formula:

$\frac{1}{2}(m_1+m_2)v^2=(m_1+m_2)gh$

Where:

m1 is the mass of the bullet

m2 is the mass of the pendulum

v is the speed

g is acceleration due to gravity

h is the height = 0.01 m

Thus, we have:

$\begin{gathered} \frac{1}{2}(0.055+4)v^2=(0.055+4)*9.8*0.01 \\ \\ 2.0275v^2=0.39739 \end{gathered}$

Solving further:

$\begin{gathered} v^2=\frac{0.39739}{2.0275} \\ \\ v^2=0.196 \\ \\ v=\sqrt{0.196} \\ \\ v=0.443\text{ m/s} \end{gathered}$

The speed of the system just after collision is 0.443 m/s.

Thus, the speed of the pendulum just after collision is 0.443 m/s.

ANSWER:

A. 0.443 m/s