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(Solved): Two blocks of masses \( m_{1}=1.83 \mathrm{~kg} \) and \( m_{2}=4.89 \mathrm{~kg} \) are each relea ...




Two blocks of masses \( m_{1}=1.83 \mathrm{~kg} \) and \( m_{2}=4.89 \mathrm{~kg} \) are each released from rest at a helght
Two blocks of masses \( m_{1}=1.83 \mathrm{~kg} \) and \( m_{2}=4.89 \mathrm{~kg} \) are each released from rest at a helght of \( \mathrm{h}=5.15 \mathrm{~m} \) on a frictionless track, as shown in the figure below, and underao an elastinihaari-an collision. Determine the velocity of the \( m_{1}=1.83 \mathrm{~kg} \) block just before the collision. \[ \text { (in } m / s \text { ) } \] Tries \( 0 / 100 \) Determine the magnitude of the velocity of the \( m_{2}=4.89 \mathrm{~kg} \) block just after the collision. (in \( m / s \) ) Tries \( 0 / 100 \) Determine the magnitude of the velocity of the \( m_{1}=1.83 \mathrm{~kg} \) block just after the collision. \[ \text { (in } \mathrm{m} / \mathrm{s}) \] Tries \( 0 / 100 \) Determine the maximum heights to which \( \mathrm{m}_{2} \) rises after collision. (in \( m \) ) Tries 0/100 Determine the maximum heights to which \( m_{1} \) rises after collision. (in m) Tries \( 0 / 100 \)


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1) velocity of m1 from the system of energy we can say that, mgh=12mv2 v=2gh=2×9.8×5.15=10.05 velocity of m1 is, v1 = 10.05 m/s similarly, v2=2gh=2×9.
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