Question

A bullet of mass 10g is fired from a rifle. The bullet tales 0.003 second to move through its barrel and leaves it with a velocity of 300m|s. What is the force exerted on the bullet by the rifle?

Answer 1

m(mass)=10g

            =10/1000kg

            =0.01kg

u(initial velocity)=0m/s(because the bullet was at rest)

v(final velocity)=300m/s

t(time)=0.003s

a(acceleration)=(v-u)/t

                       =(300-0)0.003

                       =300×1000/3

                       =300000/3

                       =100000m/s²

F(force)=ma

            =0.01×100000

            =1000N

Symbol key:

s=second

g=gram

kg=kilogram

m/s=meter per second

N=newton

Answer 2

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Correct question :

A bullet of mass 10 g is fired from a rifle. The bullet takes 0.003 second to move through its barrel and leaves it with a velocity of 300m/s. What is the force exerted on the bullet by the rifle?

◈_______________________◈

Explanation:

GiveN :

• Mass of bullet(m) = 10g = $\small{\sf{\frac{10}{1000}}}$

↦0.01kg

• Time = 0.003s
• final velocity (v) = 300m/s
• initial velocity (u) = 0

To finD:

• Force exerted by the bullet(F)

Solution:

We have given mass, time, final velocity and initial velocity, and we have to find force. hence, we will use ' Newton's second laws of motion' :

Formula using :

F = $\small{\sf{m( \frac{v - u}{t} )}}$

Where,

F = force

m = mass

v= final velocity

u = initial velocity

t = time

Solving question by putting values in given formula :

F = $\small{\sf{m( \frac{v - u}{t} )}}$

⇒F = $\small{\sf{0.01( \frac{300- 0}{0.003} )}}$

⇒F = $\small{\sf{\frac{3}{0.003} }}$

⇒F = 1000N

Hence

$\large{\boxed{\bf{F=100N}}}$