A body of mass 1 kg moving with velocity 1 m / s makes an elastic one dimensional collision with an identical stationary body. They are in contact for brief time 1 s. Their force of interaction increases from zero to F0 linearly in time 0.5 s and decreases linearly to zero in further time 0.5 s as shown in figure. The magnitude of F 0 is

Question

A body of mass 1 kg moving with velocity 1 m / s makes an elastic one dimensional collision with an identical stationary body. They are in contact for brief time 1 s. Their force of interaction increases from zero to F0 linearly in time 0.5 s and decreases linearly to zero in further time 0.5 s as shown in figure. The magnitude of F 0 is (A) 4 N (B) 1 N (C) 2 N (D) zero

Tardigrade Admin · Accepted Answer

Momentum change Δ P = vec P f- vec P i = Impulse Impulse = Area of F - t curve =(Δ P ) ⇒ (1/2) × 1 × F 0=1 ⇒ F 0=2 N

4 N

1 N

2 N

zero

Momentum change ΔP=Pf​−Pi​= Impulse Impulse = Area of F−t curve =(ΔP) ⇒21​×1×F0​=1 ⇒F0​=2N

Momentum change $Δ P = P_{f} - P_{i} =$ Impulse
Impulse $=$ Area of $F - t$ curve $= (Δ P)$
$\Rightarrow \frac{1}{2} \times 1 \times F_{0} = 1$
$\Rightarrow F_{0} = 2 N$