Two ice skaters A and B approach each other a | vedclass

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(C) The momentum of skater $A$ is $p_A = m_A v_A = 30 	imes 1 = 30 \, kg \cdot m/s$.The momentum of skater $B$ is $p_B = m_B v_B = 20 	imes 2 = 40 \

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(C) The momentum of skater $A$ is $p_A = m_A v_A = 30 	imes 1 = 30 \, kg \cdot m/s$.The momentum of skater $B$ is $p_B = m_B v_B = 20 	imes 2 = 40 \, kg \cdot m/s$.Since the skaters approach each other at right angles,the resultant momentum $p$ is given by the vector sum:$p = \sqrt{p_A^2 + p_B^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, kg \cdot m/s$.After they cling together,the total mass of the system is $M = m_A + m_B = 30 + 20 = 50 \, kg$.Using the conservation of momentum,the final velocity $v_f$ is:$v_f = \frac{p}{M} = \frac{50}{50} = 1 \, m/s$.

Two ice skaters $A$ and $B$ approach each other at right angles. Skater $A$ has a mass $30 \, kg$ and velocity $1 \, m/s$,and skater $B$ has a mass $20 \, kg$ and velocity $2 \, m/s$. They meet and cling together. Their final velocity of the couple is ............. $m/s$.

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