(C) The force on the bullet is given by $F = 600 - 2 \times 10^5 t$.When the bullet leaves the barrel,the force becomes zero,so $F = 0$.$600 - 2 \time

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) The force on the bullet is given by $F = 600 - 2 \times 10^5 t$.When the bullet leaves the barrel,the force becomes zero,so $F = 0$.$600 - 2 \times 10^5 t = 0$$2 \times 10^5 t = 600$$t = \frac{600}{2 \times 10^5} = 3 \times 10^{-3} \ s$.Impulse $I$ is defined as the integral of force with respect to time: $I = \int_{0}^{t} F \ dt$.$I = \int_{0}^{3 \times 10^{-3}} (600 - 2 \times 10^5 t) \ dt$$I = [600t - 10^5 t^2]_{0}^{3 \times 10^{-3}}$$I = 600(3 \times 10^{-3}) - 10^5(3 \times 10^{-3})^2$$I = 1.8 - 10^5(9 \times 10^{-6})$$I = 1.8 - 0.9 = 0.9 \ N-s$.

Class 11 Physics Laws of Motion Third Law of Motion and Momentum and Impulse

Difficult

$A$ bullet is fired from a gun. The force on the bullet is given by $F = 600 - 2 \times 10^5 t$,where $F$ is in newtons and $t$ is in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. What is the impulse imparted to the bullet in $N-s$?

AIPMT 1998 All AIPMT

A
$9$
B
$0$
C
$0.9$
D
$1.8$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Laws of Motion

Subtopic

Third Law of Motion and Momentum and Impulse

Previous Year

AIPMT 1998 Paper All AIPMT years →

Two billiard balls each of mass $0.05\; kg$ moving in opposite directions with speed $6\; m/s$ collide and rebound with the same speed. What is the impulse imparted to each ball due to the other (in $; Ns$)?

Easy

View Solution

$A$ particle moves in the $x-y$ plane under the action of a force $\vec F$ such that the value of its linear momentum $\vec P$ at any time $t$ is $P_x = 2 \cos t$ and $P_y = 2 \sin t$. The angle between $\vec F$ and $\vec P$ at a given time $t$ will be:

Difficult

View Solution

$A$ machine gun fires $20$ bullets per second into a target. Each bullet weighs $150 \, g$ and has a speed of $800 \, m/s$. Find the force necessary to hold the gun in position.

Easy

View Solution

$A$ particle of mass $m$ moving with velocity $(3\hat{i} + 2\hat{j}) \ m/s$ collides with another body of mass $M$ and finally moves with velocity $(-2\hat{i} + \hat{j}) \ m/s$. Then during the collision:

Medium

View Solution

$A$ body of mass $2 \, kg$ collides with a wall with a speed of $100 \, m/s$ and rebounds with the same speed. If the time of contact was $1/50 \, s$,the force exerted on the wall is:

Easy

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

$A$ bullet is fired from a gun. The force on the bullet is given by $F = 600 - 2 \times 10^5 t$,where $F$ is in newtons and $t$ is in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. What is the impulse imparted to the bullet in $N-s$?

Similar Questions

Two billiard balls each of mass $0.05\; kg$ moving in opposite directions with speed $6\; m/s$ collide and rebound with the same speed. What is the impulse imparted to each ball due to the other (in $; Ns$)?

$A$ particle moves in the $x-y$ plane under the action of a force $\vec F$ such that the value of its linear momentum $\vec P$ at any time $t$ is $P_x = 2 \cos t$ and $P_y = 2 \sin t$. The angle between $\vec F$ and $\vec P$ at a given time $t$ will be:

$A$ machine gun fires $20$ bullets per second into a target. Each bullet weighs $150 \, g$ and has a speed of $800 \, m/s$. Find the force necessary to hold the gun in position.

$A$ particle of mass $m$ moving with velocity $(3\hat{i} + 2\hat{j}) \ m/s$ collides with another body of mass $M$ and finally moves with velocity $(-2\hat{i} + \hat{j}) \ m/s$. Then during the collision:

$A$ body of mass $2 \, kg$ collides with a wall with a speed of $100 \, m/s$ and rebounds with the same speed. If the time of contact was $1/50 \, s$,the force exerted on the wall is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module