A box of mass 3 ,kg moves on a horizontal fri | vedclass

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(D) According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.$m_1

Vedclass · Accepted Answer

$72$

Vedclass · Answer

(D) According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$Given, mass of boxes $m_1 = m_2 = 3 \,kg$, speed of the moving box $u_1 = 4 \,m/s$, and initial speed of the second box $u_2 = 0$.Substituting the values:$3 	imes 4 + 3 	imes 0 = (3 + 3) v$$12 = 6v \Rightarrow v = 2 \,m/s$Thus, the two bodies move together with a velocity of $2 \,m/s$ just after the collision.Now, applying the law of conservation of energy from the edge of the table to the floor:$Total Energy_{initial} = Total Energy_{final}$$KE_{initial} + PE_{initial} = KE_{final} + PE_{final}$Taking the floor as the reference level $(h=0)$:$\frac{1}{2} (m_1 + m_2) v^2 + (m_1 + m_2) g h = KE_{final} + 0$$\frac{1}{2} 	imes (3 + 3) 	imes (2)^2 + (3 + 3) 	imes 10 	imes 1 = KE_{final}$$\frac{1}{2} 	imes 6 	imes 4 + 6 	imes 10 = KE_{final}$$12 + 60 = 72 \,J$Hence, the kinetic energy just before the boxes strike the floor is $72 \,J$.

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