Question

A bullet of mass 20g is moving with a speed of 200m/s. What is the kinetic energy of the bullet?​

Answer 1

Explanation:

Explanation:

It is given that,

Mass of a bullet, m₁ = 20 g = 0.02 kg

Speed of a bullet, v₁ = 200 m/s

The bullet strikes and get embedded in to aa stationary wooden block of mass 980g. It means that, m₂ = 980 g + 20 g = 1000 g = 1 kg

Let v₂ is the velocity with which the block moves. The momentum will remain conserved. So, using the conservation of momentum as :

So, the block will move with a velocity of 4 m/s. Hence, this is the required solution.

4m/s

Answer 2

Provided that:

• Mass of bullet = 20 g
• Velocity = 200 m/s

To calculate:

• The kinetic energy

Solution:

• The kinetic energy = 400 J

Knowledge required:

• SI unit of mass = kg
• SI unit of velocity = m/s
• SI unit of kinetic energy = J

Using concepts:

• Formula to convert g-kg
• Formula to find kinetic energy

Using formulas:

• 1 g = 1/1000 kg
• E_k = ½mv²

Where, m denotes mass, v denotes velocity and E_k denotes kinetic energy.

Required solution:

~ Firstly let us convert g-kg!

→ 1 g = 1/1000 kg

→ 20 g = 20/1000 kg

→ 20 g = 2/100 kg

→ 20 g = 0.02 kg

• Henceforth, converted!

~ Now let's calculate the kinetic energy by using suitable formula!

→ E_k = ½mv²

→ E_k = ½ · 0.02 · (200)²

→ E_k = ½ · 0.02 · 40000

→ E_k = 1 · 0.02 · 20000

→ E_k = 0.02 · 20000

→ E_k = 400 J

→ Kinetic energy = 400 Joules

• Henceforth, solved!

Know more:

➜ Kinetic energy = Energy posses by a body due to it's motion is called kinetic energy of the body. If a body of mass is moving with speed then kinetic energy of body is ${\sf{\dfrac{1}{2}mv^{2}}}$