MHT CET · Physics · Laws of Motion
Force is applied to a body of mass \(2 \mathrm{~kg}\) at rest on a frictionless horizontal surface as shown in the force against time \((\mathrm{F}-\mathrm{t})\) graph. The speed of the body after 1 second is
- A \(7.5 \mathrm{~m} / \mathrm{s}\)
- B \(12.5 \mathrm{~m} / \mathrm{s}\)
- C \(10 \mathrm{~m} / \mathrm{s}\)
- D \(15 \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(A) \(7.5 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Area under the F-t graph gives change in momentum. Since the body is initially at rest, it gives the momentum of the body after 1 second.
\(\begin{aligned}& \text { Area }=10 \times 0.5+20 \times 0.5=5+10=15 \mathrm{~N}-\mathrm{s} \\& \therefore \mathrm{mV}=15 \mathrm{~N}-\mathrm{s} \\& \mathrm{V}=\frac{15}{\mathrm{~m}}=\frac{15}{2}=7.5 \mathrm{~m} / \mathrm{s}\end{aligned}\)
\(\begin{aligned}& \text { Area }=10 \times 0.5+20 \times 0.5=5+10=15 \mathrm{~N}-\mathrm{s} \\& \therefore \mathrm{mV}=15 \mathrm{~N}-\mathrm{s} \\& \mathrm{V}=\frac{15}{\mathrm{~m}}=\frac{15}{2}=7.5 \mathrm{~m} / \mathrm{s}\end{aligned}\)
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