A particle of mass $m$ moving with velocity $u$ 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 the figure. The magnitude of $F_{0}$ is

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

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

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

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Solution

The correct option is A $\frac{2 m u}{t_{0}}$
Change in momentum $=$ Impulse $=$ Area under $F - t$ curve.

$| Δ p | = \int_{0}^{t_{0}} F_{a v g} d t$ ;

Area of the triangle in $F - t$ plot is $= \frac{1}{2} \times b a s e \times h e i g h t = \frac{1}{2} t_{0} F_{0}$

In elastic collision the velocities are exchanged if both particles have same mass.

$| Δ p | = m u$

$∴ m u = \frac{1}{2} t_{0} F_{0}$

$\Rightarrow F_{0} = \frac{2 m u}{t_{0}}$

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The correct option is A 2mut0Change in momentum = Impulse = Area under F−t curve.|Δp|=∫t00Favgdt;Area of the triangle in F−t plot is =12×base×height=12t0F0In elastic collision the velocities are exchanged if both particles have same mass.|Δp|=mu∴mu=12t0F0⇒F0=2mut0