Question

A body of $2kg$ has an initial speed $5m{s}^{-1}$. A force acts on it for $6.5$ seconds in the direction of motion. The force time graph is shown in figure. The final speed of the body is

• A. $9.25m{s}^{-1}$
• B. $5m{s}^{-1}$
• C. $14.31m{s}^{-1}$
• D. $4.25m{s}^{-1}$

Answer 1

Answer: (C)

$14.31m{s}^{-1}$

We know,
Area under force-time graph $=$ change in linear momentum
$\therefore \Delta P=\frac{1}{2}\times 2\times 4+(4-2)\times 4+(4.5-4)\times 2.5+\frac{1}{2}\times (4.5-4)(4-2.5)+(6.5-4.5)\times 2.5$
$\therefore P_f-P_i=18.625$
Now,
$P_i=m\times v_i=2\times 5=10kgm/s$
$\therefore P_f=10+18.625$
$\therefore P_f=28.625kgm{s}^{-1}$
$\therefore v_f=14.3125m/s$ (Since $m=2kg$)