A body of mass 4 kg collides head-on elastica | vedclass

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(B) The impulse $J$ acting on the body is given by the product of the average force $F$ and the time duration $\Delta t$ of the collision.$J = F 	ime

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$1$

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(B) The impulse $J$ acting on the body is given by the product of the average force $F$ and the time duration $\Delta t$ of the collision.$J = F 	imes \Delta t$Given $F = 100\ N$ and $\Delta t = 0.02\ s$,the impulse is:$J = 100\ N 	imes 0.02\ s = 2\ N\cdot s$According to the impulse-momentum theorem,the impulse applied to a body is equal to the change in its linear momentum.$J = \Delta p = m_2(v_2 - u_2)$Since the $2\ kg$ body was initially at rest,$u_2 = 0$. Thus:$2 = 2\ kg 	imes (v_2 - 0)$$2 = 2 	imes v_2$$v_2 = 1\ m/s$Therefore,the velocity of the $2\ kg$ body after the impact is $1\ m/s$.

$A$ body of mass $4\ kg$ collides head-on elastically with another body of mass $2\ kg$ kept at rest in free space. The time of collision is $0.02\ s$ and the average impulsive force acting on each body is $100\ N$. Find the velocity of the $2\ kg$ body after the impact. (in $m/s$)

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