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  4. A particle of mass m moving with velocity u makes an elastic one dimensional collision with a stationary particle of mass m. They are in contact for a short time T. Their force of interaction increases from zero to F0 linearly in time (T/2) and decreases linearly to zero in further time (T/2) (shown in figure). The magnitude of F0 is

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Q. A particle of mass $m$ moving with velocity $u$ makes an elastic one dimensional collision with a stationary particle of mass $m$. They are in contact for a short time $T$. Their force of interaction increases from zero to $F_{0}$ linearly in time $\frac{T}{2}$ and decreases linearly to zero in further time $\frac{T}{2}$ (shown in figure). The magnitude of $F_{0}$ isPhysics Question Image

Bihar CECEBihar CECE 2012Laws of Motion

Solution:

In elastic one dimensional collision particle rebounds with same speed in opposite direction.
So, change in momentum $=2\, mu$
But, impulse $=F \times T=$ Change in momentum
$\Rightarrow F_{0} \times T=2\, m u$
or $F_{0}=\frac{2\, m u}{T}$