A particle of mass m moving with a velocity u | vedclass

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Question

(C) In an elastic one-dimensional collision between two particles of equal mass $m$, the particles exchange their velocities. Since the first particle

Vedclass · Accepted Answer

$\frac{4mu}{3T}$

Vedclass · Answer

(C) In an elastic one-dimensional collision between two particles of equal mass $m$, the particles exchange their velocities. Since the first particle was moving with velocity $u$ and the second was stationary, after the collision, the first particle comes to rest and the second particle moves with velocity $u$.The change in momentum of the second particle is $\Delta p = m(u - 0) = mu$.According to the impulse-momentum theorem, the impulse $J$ is equal to the change in momentum $\Delta p$. The impulse is also equal to the area under the force-time $(F-t)$ graph.The area of the trapezium formed by the graph is:$J = 	ext{Area} = \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$$J = \frac{1}{2} 	imes (T + \frac{T}{2}) 	imes F_0$$J = \frac{1}{2} 	imes (\frac{3T}{2}) 	imes F_0 = \frac{3T F_0}{4}$Equating impulse to the change in momentum:$mu = \frac{3T F_0}{4}$$F_0 = \frac{4mu}{3T}$

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