
A machine gun fires a bullet of mass 65g with a velocity of 1300m/s. The man holding it can exert a maximum force of 169N on the gun. The number of bullets he can fire per second will be
A. 1
B. 2
C. 3
D. 4
Answer
124.2k+ views
Hint: As we have to find out the number of bullets he can fire per second. So we can use here the formula of impulse which is equal to the product of change of momentum. On substituting the values we can find easily the value of the number of bullets he can fire per second.
Complete step-by-step answer:
It is given that mass of bullet is $M = 65kg = 0.065g$
Initially the bullet was it at rest so its initial velocity is $U = 0$
Machine fires the bullet with velocity of 1300m/s, therefore
Final velocity of bullet is $V = 1300m/s$
Force exerted on gun $F = 169N$
Now, as we know that impulse is equal to the change of momentum, therefore we will find change in momentum first. As change is momentum is the difference of final momentum and the initial momentum, therefore
Change in momentum $P = MV - MU = M\left( {V - U} \right)$
For N bullets the change in momentum is $\Delta P = NM\left( {V - U} \right)$
Now Impulse= change in momentum that is $I = \Delta P$…………………………(1)
And we also know that impulse is the product of force and time. So $I = F \times T$
Put this value in equation (1)
⇒ $F \times T = NM\left( {V - U} \right)$
Now substituting the values we get,
⇒$168 \times 1 = N \times 0.065\left( {1300 - 0} \right)$
⇒ $N = \dfrac{{169}}{{0.065 \times 1300}} = 2$
⇒ Number of bullets fire per second = 2
Hence, B option is correct.
Note: Impulse is the integral of the force over the time interval for which it acts. This is also a vector quantity like force. Impulse applied to any object produces an equivalent vector change in its linear momentum. Here, we need to find the number of bullets fired per second so when we write impulse as the product of force and time so we must take time equal to one second.
Complete step-by-step answer:
It is given that mass of bullet is $M = 65kg = 0.065g$
Initially the bullet was it at rest so its initial velocity is $U = 0$
Machine fires the bullet with velocity of 1300m/s, therefore
Final velocity of bullet is $V = 1300m/s$
Force exerted on gun $F = 169N$
Now, as we know that impulse is equal to the change of momentum, therefore we will find change in momentum first. As change is momentum is the difference of final momentum and the initial momentum, therefore
Change in momentum $P = MV - MU = M\left( {V - U} \right)$
For N bullets the change in momentum is $\Delta P = NM\left( {V - U} \right)$
Now Impulse= change in momentum that is $I = \Delta P$…………………………(1)
And we also know that impulse is the product of force and time. So $I = F \times T$
Put this value in equation (1)
⇒ $F \times T = NM\left( {V - U} \right)$
Now substituting the values we get,
⇒$168 \times 1 = N \times 0.065\left( {1300 - 0} \right)$
⇒ $N = \dfrac{{169}}{{0.065 \times 1300}} = 2$
⇒ Number of bullets fire per second = 2
Hence, B option is correct.
Note: Impulse is the integral of the force over the time interval for which it acts. This is also a vector quantity like force. Impulse applied to any object produces an equivalent vector change in its linear momentum. Here, we need to find the number of bullets fired per second so when we write impulse as the product of force and time so we must take time equal to one second.
Recently Updated Pages
Difference Between Circuit Switching and Packet Switching

Difference Between Mass and Weight

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main 2025 Helpline Numbers - Center Contact, Phone Number, Address

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main Login 2045: Step-by-Step Instructions and Details

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
