Question

If a bullet of mass of mass 50 gram is moving with a velocity of 200
km/h, calculate the kinetic energy stored in it.

Answer 1

Given :

• Mass of the bullet = 50 g

• Velocity of the bullet = 200 km/h

To find :

• Kinetic energy stored in the bullet

Solution :

Kinetic energy of a body is given by ,

$\dag \boxed {\rm{KE = \: \frac{1}{2}m {v}^{2} }}$

Where ,

• m is mass of the body
• v is velocity of the body

The relation between kg and g is

$: \implies {\rm{1kg= 10000 \: g}} \\ \\ : \implies \rm \: 1g = \frac{1}{1000} \: kg \\ \\ : \implies \boxed {\rm{ 1g = 10 {}^{ - 3} \: kg }}$

Then , 50 g = 50 × 10⁻³ kg = 5 × 10⁻² kg

Now , the relation between m/s and km/hr is ,

$\boxed {\rm{1km. {hr}^{ - 1} = 0.278 \: m {s}^{ - 1} }}$

Then , 200 km/hr = 200 × 0.278 m/s = 55.6 m/s

We have ,

• m = 50 g = 5 × 10⁻² kg
• v = 200 km/hr = 55.6 m/s

$: \implies \rm \: KE = \dfrac{1}{2}(5 \times {10}^{ - 2} \: kg)(55.6 \: m {s}^{ - 1} ) {}^{2} \\ \\ : \implies \rm \: KE = \frac{1}{2} (5\times {10}^{ - 2} \: kg)({ 3091.36} \: m {}^{2} {s}^{ - 2} ) \\ \\ : \implies \rm \: KE = \frac{1}{2} (15456.8 \times {10}^{ - 2} \: kgm {}^{2} {s}^{ - 2} ) \\ \\ : \implies \rm \:KE = 7728.4 \times {10}^{ - 2} \: \: J \\ \\ : \implies \rm \: KE = 77.284 \: J$

∴ The kinetic energy stored in the bullet of mass 50 g and velocity 200 m/s is 77.284 J

Answer 2

Answer:

Given :

Mass of the bullet = 50 g

Velocity of the bullet = 200 km/h

To find :

Kinetic energy stored in the bullet

Solution :

Kinetic energy of a body is given by ,

\dag \boxed {\rm{KE = \: \frac{1}{2}m {v}^{2} }}†

KE=

2

1

mv

2

Where ,

m is mass of the body

v is velocity of the body

The relation between kg and g is

\begin{gathered} : \implies {\rm{1kg= 10000 \: g}} \\ \\ : \implies \rm \: 1g = \frac{1}{1000} \: kg \\ \\ : \implies \boxed {\rm{ 1g = 10 {}^{ - 3} \: kg }}\end{gathered}

:⟹1kg=10000g

:⟹1g=

1000

1

kg

:⟹

1g=10

−3

kg

Then , 50 g = 50 × 10⁻³ kg = 5 × 10⁻² kg

Now , the relation between m/s and km/hr is ,

\boxed {\rm{1km. {hr}^{ - 1} = 0.278 \: m {s}^{ - 1} }}

1km.hr

−1

=0.278ms

−1

Then , 200 km/hr = 200 × 0.278 m/s = 55.6 m/s

We have ,

m = 50 g = 5 × 10⁻² kg

v = 200 km/hr = 55.6 m/s

\begin{gathered} : \implies \rm \: KE = \dfrac{1}{2}(5 \times {10}^{ - 2} \: kg)(55.6 \: m {s}^{ - 1} ) {}^{2} \\ \\ : \implies \rm \: KE = \frac{1}{2} (5\times {10}^{ - 2} \: kg)({ 3091.36} \: m {}^{2} {s}^{ - 2} ) \\ \\ : \implies \rm \: KE = \frac{1}{2} (15456.8 \times {10}^{ - 2} \: kgm {}^{2} {s}^{ - 2} ) \\ \\ : \implies \rm \:KE = 7728.4 \times {10}^{ - 2} \: \: J \\ \\ : \implies \rm \: KE = 77.284 \: J \end{gathered}

:⟹KE=

2

1

(5×10

−2

kg)(55.6ms

−1

)

2

:⟹KE=

2

1

(5×10

−2

kg)(3091.36m

2

s

−2

)

:⟹KE=

2

1

(15456.8×10

−2

kgm

2

s

−2

)

:⟹KE=7728.4×10

−2

J

:⟹KE=77.284J

∴ The kinetic energy stored in the bullet of mass 50 g and velocity 200 m/s is 77.284 J