A particle of mass $m$ moving with velocity $v$ makes an elastic one–dimensional collision with a stationary particle of mass $m$ They come in contact for a very small time $t_{0} .$ Their force of interaction increases from zero to $F_{0}$ linearly in time $0.5 t_{0},$ and decreases linearly to zero in further time $0.5 t_{0}$ as shown in figure. The magnitude of $F_{0}$ is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$\frac{2 m u}{t_{0}}$

$\frac{m u}{2 t_{0}}$

$\frac{m u}{t_{0}}$

None of these

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

In elastic collision between equal masses velocities are interchanged. Therefore, change in momentum in any one particle is $m u .$
Now, $| Δ p | = | i m p u l s e |$
$= a r e a u n d e r F - t g r a p h ∴ m u = \frac{1}{2} \times t_{0} \times F_{0} ∴ F_{0} = \frac{2 m u}{t_{0}}$