Question

what will be the kinetic energy of a body when its mass is made 4 times and velocity is doubled??answer is 16 times to the original value
but how please explain me please tell me because it's very urgent​

Answer 1

Given:-

• The mass is made 4 times.

• The Velocity is Doubled.

To Find:-

• The Kinetic energy of the Body.

Formulae used:-

• K. E = ½ × mv²

Where,

• K.E = Kinetic Energy
• m = Mass
• v = Velocity

ATQ,

• m = 4m

• v = 2m.

Now,

→ Original K. E ( x ) = ½ × mv²...........1

Therefore,

→ New K. E ( y ) = ½ × 4 × m × 2 × v²

→ K. E = ½ × 4m × (2v)²

→ K. E = ½ × 4m × 4v²

→ K. E = 4 × 4 ( ½ × mv²)

→ K. E = 16( ½ × mv² )

Using eq. 1

→ y = 16x

Hence, The New kinetic energy is 16 times to that of original.

Answer 2

Given:

• Final Mass = 4 × Initial mass
• Final velocity = 2× Initial velocity

━━━━━━━━━━━━━━━━

Need to find:

• The Final Kinetic energy =?

━━━━━━━━━━━━━━━━

Soluton:

Let initial mass = m1

Then final mass = m2 = 4m1

Let initial velocity be = v1

Then Final velocity = v2 = 2v1

━━━━━━━━━━

Now initial kinetic energy:

$= \dfrac{1}{2} {m}_{1} \: {{v}_{1}}^{2}$

━━━━━━━━━━

Now final Kinetic energy:

$= \dfrac{1}{2} {m}_{2} {{v}_{2}}^{2}$

$= \frac{1}{2} \times (4{m}_{1}) \times {(2{v}_{1})}^{2}$

$= \dfrac{1}{2} \times 4{m}_{1} \times 4 {v}_{1}^{2}$

$= 4\times4 \times \dfrac{1}{2} \times{m}_{1} {v}_{1}^{2}$

$\red{\bold{\boxed{\large{\boxed{= 16 \times {Kinetic\:Energy} _{ initial}}}}}}$