MHT CET · Physics · Laws of Motion
A body of mass \(2 \mathrm{~kg}\) collides with a wall with a speed of \(100 \mathrm{~ms}^{-1}\) and rebounds with the same speed. If the time of contact is \(1 / 50 \mathrm{~s}\), the force exerted on the wall is
- A \(8 \mathrm{~N}\)
- B \(2 \times 10^{4} \mathrm{~N}\)
- C \(4 \mathrm{~N}\)
- D \(10^{4} \mathrm{~N}\)
Answer & Solution
Correct Answer
(B) \(2 \times 10^{4} \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
We know that, \(F=\frac{d p}{d t}=\frac{m d v}{d t}=\frac{m \times 2 v}{1 / 50}\)
\(=\frac{2 \times 2 \times 100}{1 / 50}=2 \times 10^{4} \mathrm{~N}\)
\(=\frac{2 \times 2 \times 100}{1 / 50}=2 \times 10^{4} \mathrm{~N}\)
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