Question

a bullet of mass 50g moving with a speed of 500 m/s penetrates a fixed target which offers a constant resistive force of 1000N to the motion of the bullet. Find initial kinetic energy of the bullet and the distance through which the bullet has penetrated. ​

Answer 1

Answer:

Given,

mass=5g

speed=500m/s

force=1000N

(a) kinetic energy=

2

1

×m×v×v

m=5g=

1000

5

kg

v=500

⇒

2

1

×

1000

5

×500×500

=625J

(b) force=mass×acceleration

1000=

1000

5

×acceleration

Acceleration=1000×

5

1000

=200000m/s

2

using the third equation of motion

v

2

=u

2

+2as

(500)

2

=0+2×200000×s

250000=400000×s

s=0.625m

Answer 2

Answer:

mass=50g

$$speed=500m/s$$

$$force=1000N$$

(a) kinetic energy=\dfrac{1}{2}\times m\times v\times

$$m=5g=\dfrac{5}{1000}kg$$

$$v=500$$

\Rightarrow \dfrac{1}{2}\times\dfrac{5}{1000}\times500\times500

=$$625J$$

(b) force= mass\times acceleration

1000=\dfrac{50}{1000}\times acceleration

Acceleration=1000\times \dfrac{1000}{5}

=200000m/s^2

using the third equation of motion

v^2=u^2 +2as

(500)^2=0+2\times 200000\times s

250000=400000\times s

s=625m