A force of 1 N acts upon a mass of 1 kg for 1 second and gives it a momentum $P_{1}$ and kinetic energy $E_{1}$ . The same force acts upon the same mass for a displacement of 1 m and gives it a momentum $P_{2}$ and kinetic energy $E_{2}$ . Then, which of the following is correct?

P_{1} = P_{2}, E_{1} = E_{2}

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P_{1} > P_{2}, E_{1} > E_{2}

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P_{1} < P_{2}, E_{1} < E_{2}

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P_{1} > P_{2}, E_{1} < E_{2}

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Solution

The correct option is D $P_{1} < P_{2}, E_{1} < E_{2}$
$P_{1} = F t = 1$ N $\times 1$ s $= 1$ Ns
$a = \frac{F}{m}$
$a = \frac{1}{1} = 1$ m/s $^{2}$
$v = u + a t$
$v = 0 + (1) (1) = 1$ m/s $^{2}$
$E_{1} = \frac{1}{2} m v^{2}$
$E_{1} = \frac{1}{2} \times 1 \times (a t)^{2}$
$E_{1} = \frac{1}{2} \times 1 \times 1 = 0.5$ J
$s = u t + \frac{1}{2} a t^{2}$
$\Rightarrow 1 = (0) t + \frac{1}{2} (1) t^{2}$
$\Rightarrow t = \sqrt{2}$ s
$v = u + a t$
$v = 0 + (1) \sqrt{2} = \sqrt{2}$ m/s
$P_{2} = m v = (1) (\sqrt{2}) = \sqrt{2}$ kgm/s
$E_{2} = \frac{1}{2} m v^{2}$
$\Rightarrow E_{2} = \frac{1}{2} \times 1 \times (\sqrt{2})^{2} = 1$ J
$P_{1} < P_{2}$ and $E_{1} < E_{2}$

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A force of 1 N acts upon a mass of 1 kg for 1 second and gives it a momentum P1 and kinetic energy E1. The same force acts upon the same mass for a displacement of 1 m and gives it a momentum P2 and kinetic energy E2. Then, which of the following is correct?

The correct option is D P1<P2,E1<E2P1=Ft=1N×1s=1Nsa=Fma=11=1m/s2v=u+atv=0+(1)(1)=1m/s2E1=12mv2E1=12×1×(at)2E1=12×1×1=0.5 Js=ut+12at2⇒1=(0)t+12(1)t2⇒t=√2 sv=u+atv=0+(1)√2=√2 m/sP2=mv=(1)(√2)=√2 kgm/sE2=12mv2⇒E2=12×1×(√2)2=1 JP1<P2 and E1<E2

A force of 1 N acts upon a mass of 1 kg for 1 second and gives it a momentum $P_{1}$ and kinetic energy $E_{1}$ . The same force acts upon the same mass for a displacement of 1 m and gives it a momentum $P_{2}$ and kinetic energy $E_{2}$ . Then, which of the following is correct?