Answer to Question #135419 in Mechanics | Relativity for Bawe

Question

A bullet of mass 25 g and travelling horizontally at a speed of 200 m/s imbeds itself in a wooden block of mass 5 kg suspended by cords 3 m long. How far will the block swing from its position of rest before beginning to return?

Answer 1

Let A be the initial position of the block

Let B be the final position of the block before it starts returning

Let m1 be the mass of the bullet

Let u1 be the initial speed of the bullet

Let m2 be the mass of the block

Let u2 be the initial speed of the block

Let v be the final speed of the combined mass

Applying the principle of conservation of momentum

the total momentum before collision is equal to the total momentum after collision.

$$m1u1 + m2u2 = (m1 + m2)v\\ 0.025 \times 200 = (0.025 + 5)v\\ v = 0.995 m/s\\$$

The initial kinetic energy of combined mass after collision

$$=\frac{1}{2} \times (m1+m2) × v^2\\ =\frac{1}{2} × (0.025 + 5) × 0.995\\ = 2.487J\\$$

the potential energy of the combined mass before beginning to return

$$=(m1 + m2) × g × h\\ = 5.025 × 9.81 × h$$

Applying the principle of conservation of energy

the initial kinetic energy of the combined mass after collision is equal to the potential energy of the combined mass before beginning to return

$$2.487J = 5.025 × 9.81 × h\\ h = 0.05m$$

from the diagram above,

$$h=|AC|\\ |OC|=|OA|-|CA|\\ |OC|= (3-0.05)m= 2.95m\\ |CB|= \sqrt{|OB|²-|OC|} = \sqrt{3²-2.95²}= 0.545m\\$$

The distance between the final and initial position is represented by |AB|

$|AB|=\sqrt{|AC|²+|CB|²}\\ |AB|=\sqrt{0.05²+0.545²}\\ |AB|=0.547m$