Question

Force-time (F-t) graph for a particle of mass 2 kg shown in the figure. If initial velocity of the particle is –5 m/s, then calculate velocity at t = 0.5 s.



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Answer 1

Given:

Force-time (F-t) graph for a particle of mass 2 kg shown in the figure. If initial velocity of the particle is –5 m/s

To find:

Velocity at t = 0.5 sec .

Calculation:

Force is often defined as the instantaneous rate of change of momentum with respect to time.

So, area under the force-time graph will give us change in momentum;

$\sf{ \therefore \: area \: of \: \Delta = change \: in \: momentum}$

$\sf{ = > \: area \: of \: \Delta =mv - mu}$

$\sf{ = > \: area \: of \: \Delta =m(v - u)}$

$\sf{ = > \: \dfrac{1}{2} \times base \times height =m(v - u)}$

$\sf{ = > \: \dfrac{1}{2} \times 0.5 \times 200 =2(v - u)}$

$\sf{ = > \: 50 =2(v - u)}$

$\sf{ = > \: 25 =(v - u)}$

$\sf{ = > \: v - ( - 5) = 25}$

$\sf{ = > \: v + 5 = 25}$

$\sf{ = > \: v = 20 \: m {s}^{ - 1} }$

So, final answer is:

$\boxed{ \rm{ \: v = 20 \: m {s}^{ - 1} }}$

Answer 2

We need to use the property that, area under force - time graph gives impulse.

Impulse is ΔP = m * Δv or we can say,

Impulse = Force * time

Here,we have to use Impulse = ΔP = m * Δv

→ m(v - u)

So, we know,

Area under force - time graph = m(v - u)

Finding Area under graph (till t = 0.5s) :-

(1/2) * base * height

→ (1/2) * 0.5 * 200

→ 50

So,

50 = m(v - u)

It is given that m = 2kg and u = -5m/s

So,

→ 50 = 2 * [v - (-5)]

→ 50 = 2 * [v + 5]

→ 25 = v + 5

→ v = 20 m/s

Ans. v = 20 m/s