TS EAMCET · Physics · Work Power Energy
A box of mass \(3 \mathrm{~kg}\) moves on a horizontal frictionless table and collides with another box of mass \(3 \mathrm{~kg}\) initially at rest on the edge of the table at height \(1 \mathrm{~m}\). The speed of the moving box just before the collision is \(4 \mathrm{~m} / \mathrm{s}\). The two boxes stick together and fall from the table. The kinetic energy just before the boxes strike the floor is (Assume, acceleration due to gravity, \(g=10 \mathrm{~m} / \mathrm{s}^2\) )
- A \(40 \mathrm{~J}\)
- B \(80 \mathrm{~J}\)
- C \(96 \mathrm{~J}\)
- D \(72 \mathrm{~J}\)
Answer & Solution
Correct Answer
(D) \(72 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
\( \text { The given situation can be shown as below, } \) According to law of conservation of momentum, total momentum before collision \(=\) total momentum after collision. \( m_1 u_1+m_2 u_2=\left(m_1+m_2\right) v \) Given, mass of boxes, \(\quad m_1=m_2=3 \mathrm{~kg}\)…
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