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A girl of mass 40 kg jumps with horizontal velocity of 6 m/s onto a stationary cart with frictionless…

By the law of conservation of linear momentum, \(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \implies 40 \times 6 + 4 \times 0 = (40 + 4) v \implies v = \frac{240}{44} = \frac{60}{11}…

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A girl of mass 40 kg jumps with horizontal velocity of 6 m/s onto a stationary cart with frictionless wheels. If the mass of cart is 4 kg, then the final velocity of system will be (No external force is working in horizontal direction) -

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By the law of conservation of linear momentum, m1u1+m2u2=(m1+m2)v  ⟹  40×6+4×0=(40+4)v  ⟹  v=24044=6011≈5.45 m/sm_1 u_1 + m_2 u_2 = (m_1 + m_2) v \implies 40 \times 6 + 4 \times 0 = (40 + 4) v \implies v = \frac{240}{44} = \frac{60}{11} \approx 5.45\text{ m/s}.

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