A man of mass 70 kg jumps to a height of 0.8 m from the ground, then the momentum transferred by the ground to the man is
(g = 10 m/s\(^{-2}\)):
A man of mass 70 kg jumps to a height of 0.8 m from the ground, then the momentum transferred by the ground to the man is
(g = 10 m/s\(^{-2}\)):
Show Hint
- \(280 { kg ms}^{-1}\)
- \(200 { kg ms}^{-1}\)
- \(560 { kg ms}^{-1}\)
- \(400 { kg ms}^{-1}\)
The Correct Option is A
Solution and Explanation
Step 1: Calculate the initial velocity required to reach 0.8 m.
Using the formula \( v^2 = u^2 + 2as \) and rearranging for \( u \) when \( v = 0 \), \( a = -g \), and \( s = 0.8 { m} \):
\[ 0 = u^2 - 2 \times 10 \times 0.8 \Rightarrow u^2 = 16 \Rightarrow u = 4 { m/s}. \] Step 2: Calculate the momentum transferred.
\[ p = m \times u = 70 \times 4 = 280 { kg ms}^{-1}. \] The momentum calculation involves the mass and initial velocity, assuming no air resistance and perfect energy conversion.
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