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?
Answers
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
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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?
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Explanation:
GiveN :
- Mass of bullet(m) = 10g =
↦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 =
Where,
F = force
m = mass
v= final velocity
u = initial velocity
t = time
Solving question by putting values in given formula :
F =
⇒F =
⇒F =
⇒F = 1000N
Hence