A body of mass 2 , kg moving with a velocity | vedclass

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(C) According to the law of conservation of linear momentum,the total momentum before the collision is equal to the total momentum after the collision

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$2/3$

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(C) According to the law of conservation of linear momentum,the total momentum before the collision is equal to the total momentum after the collision.Let $m_1 = 2 \, kg$,$v_1 = 3 \, m/s$,$m_2 = 1 \, kg$,and $v_2 = -4 \, m/s$ (since it is moving in the opposite direction).Let the common velocity after the collision be $v$.Since the bodies stick together,the final mass is $(m_1 + m_2)$.The equation is: $m_1 v_1 + m_2 v_2 = (m_1 + m_2)v$.Substituting the values: $(2 	imes 3) + (1 	imes -4) = (2 + 1)v$.$6 - 4 = 3v$.$2 = 3v$.$v = 2/3 \, m/s$.

$A$ body of mass $2 \, kg$ moving with a velocity of $3 \, m/s$ collides head-on with a body of mass $1 \, kg$ moving in the opposite direction with a velocity of $4 \, m/s$. After the collision,the two bodies stick together and move with a common velocity. What is the common velocity in $m/s$?

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