SATHEE: Centre of Mass 2 Question 3

Centre of Mass 2 Question 3

4. A particle of mass $m$ is moving in a straight line with momentum $p$ . Starting at time $t = 0$ , a force $F = k t$ acts in the same direction on the moving particle during time interval $T$ , so that its momentum changes from $p$ to $3 p$ . Here, $k$ is a constant. The value of $T$ is (JEE Main 2019, 11 Jan Shift II)

(a) $\sqrt{\frac{2 p}{k}}$

(b) $2 \sqrt{\frac{p}{k}}$

(d) $2 \sqrt{\frac{k}{p}}$

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Answer:

Correct Answer: 4. (b)

Solution:

Here,

$F = k t$

When $t = 0$ , linear momentum $= p$

When $t = T$ , linear momentum $= 3 p$

According to Newton’s second law of motion,

applied force, $F = \frac{d p}{d t}$

or $d p = F \cdot d t$

or $d p = k t \cdot d t$

Now, integrate both side with proper limit

$\begin{aligned} \int_{p}^{3 p} d p = k \int_{0}^{T} t d t or [p]_{p}^{3 p} = k \frac{t^{2}}{2} \\ or (3 p - p) = \frac{1}{2} k [T^{2} - 0] \\ or T^{2} = \frac{4 p}{k} or T = 2 \sqrt{\frac{p}{k}} \end{aligned}$

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Centre of Mass 2 Question 3

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Centre of Mass 2 Question 3

4. A particle of mass m is moving in a straight line with momentum p. Starting at time t=0, a force F=kt acts in the same direction on the moving particle during time interval T, so that its momentum changes from p to 3p. Here, k is a constant. The value of T is (JEE Main 2019, 11 Jan Shift II)

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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