Question

A particle of mass 5kg moves in a straight line under the influence of a momentum which is varying parabolically, If the rate of change of momentum at t = 0 is 80 N. After 2 second rate of change of momentum is 480 N. What will be the velocity of particle after 5 second when it started from rest from origin ?

• A. 580 m/s
• B. 330 m/s
• C. 680 m/s
• D. 250 m/s

Answer 1

The correct option is A 580 m/s

$\text{Let momentum,}P=a{t}^2+bt+c$

$\text{force}F=\frac{dP}{dt}\text{at}t=0\text{is}80N$

$F=2a\times 0+b$

$80=b$

$\Rightarrow b=80$

$\text{Now , at t}=2\text{sec}$

$F=2a\times 2+80$

$\frac{480-80}{4}=a$

$a=100$

$F=2at+80$

$\text{Now}F=200t+80$

$\text{Acceleration}=\frac{F}{m}=\frac{200t+80}{5}=40t+16$

$\text{Acceleration}=\frac{dv}{dt}$

$dv=(40t+16)dt$

${\int }_0^{\upsilon }d\upsilon ={\int }_0^5(40t+16)dt$

$\upsilon -0={[40\times \frac{t^2}{2}+16t]}_0^5$

$\upsilon =40\times \frac{5^2}{2}+16\times 5$

$\upsilon =25\times 20+80$

$\upsilon =580m/s$