A body of mass 2 kg is initially at rest. It starts moving unidir… - Solution | JEE Physics

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A body of mass $2 \, \text{kg}$ is initially at rest. It starts moving unidirectionally under the influence of a source of constant power $P$ . Its displacement in $4 \, \text{s}$ is $\frac{1}{3} \alpha^2 \sqrt{P}$ meters. The value of $\alpha$ will be _____

Answer

Correct answer:4

Step-by-step solution

Standard Method

Given: mass $m = 2 \, \text{kg}$ , initial velocity $v_0 = 0$ , constant power $P$ , and time $t = 4 \, \text{s}$ .

Find: the value of $\alpha$ from the displacement expression.

For constant power,

\frac{1}{2}mv^2 = Pt

So,

v = \sqrt{\frac{2Pt}{m}}

Using $v = \frac{dx}{dt}$ ,

\frac{dx}{dt} = \sqrt{\frac{2Pt}{m}}

Now integrate with respect to time:

x = \sqrt{\frac{2P}{m}} \int t^{1/2} \, dt

x = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3} t^{3/2}

At $m = 2 \, \text{kg}$ and $t = 4 \, \text{s}$ ,

x = \sqrt{P} \cdot \frac{2}{3} \cdot 4^{3/2}

x = \frac{16}{3} \sqrt{P}

Given,

x = \frac{1}{3} \alpha^2 \sqrt{P}

Comparing,

\alpha^2 = 16

\alpha = 4

Therefore, the value of $\alpha$ is $4$ .

Comparison Step

From the derived displacement,

x = \frac{16}{3} \sqrt{P}

and from the question,

x = \frac{1}{3} \alpha^2 \sqrt{P}

Equating the coefficients of $\sqrt{P}$ ,

\frac{1}{3} \alpha^2 = \frac{16}{3}

\alpha^2 = 16

Hence,

\alpha = 4

So the required numerical value is $4$ .

Common mistakes

Using $P = Fv$ and treating force as constant is incorrect. The power is constant here, so as velocity changes, force must change. Start from energy gained in time $t$ : $Pt = \frac{1}{2}mv^2$ .
Writing $v = \sqrt{\frac{m}{2Pt}}$ is the inverse of the correct relation. From $\frac{1}{2}mv^2 = Pt$ , the correct result is $v = \sqrt{\frac{2Pt}{m}}$ .
Forgetting that $v = \frac{dx}{dt}$ leads to using velocity directly as displacement. After finding $v(t)$ , integrate with respect to time to obtain $x(t)$ .

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