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(B) In an elastic head-on collision between two particles of equal mass,the particles exchange their velocities. Therefore,the velocity of the inciden

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$\frac{4 m v}{3 T}$

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(B) In an elastic head-on collision between two particles of equal mass,the particles exchange their velocities. Therefore,the velocity of the incident particle becomes $0$ and the stationary particle moves with velocity $v$.The change in momentum of the incident particle is $\Delta p = m(v - 0) = mv$.According to the impulse-momentum theorem,the impulse (area under the $F-t$ graph) is equal to the change in momentum.$	ext{Area} = \Delta p = mv$.The area under the given trapezoidal $F-t$ graph is:$	ext{Area} = \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$$	ext{Area} = \frac{1}{2} 	imes \left( T + \left( \frac{3T}{4} - \frac{T}{4} ight) ight) 	imes F_0$$	ext{Area} = \frac{1}{2} 	imes \left( T + \frac{2T}{4} ight) 	imes F_0 = \frac{1}{2} 	imes \left( T + \frac{T}{2} ight) 	imes F_0 = \frac{1}{2} 	imes \left( \frac{3T}{2} ight) 	imes F_0 = \frac{3T F_0}{4}$.Equating the area to the change in momentum:$\frac{3T F_0}{4} = mv$$F_0 = \frac{4mv}{3T}$.

$A$ particle of mass $m$,moving with a velocity $v$,makes an elastic collision in one dimension with a stationary particle of mass $m$. During the collision,they remain in contact with each other for an extremely small time $T$. Their force of contact with time is shown in the figure. Then $F_0 =$

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