SATHEE: Laws of Motion - Result Question 2

Laws of Motion - Result Question 2

2. The force ’ $F$ ’ acting on a particle of mass ’ $m$ ’ is indicated by the force-time graph shown below. The change in momentum of the particle over the time interval from zero to $8 s$ is :

[2014]

(a) $24 N s$

(b) $20 N s$

(c) $12 N s$

(d) $6 N s$

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Answer:

Correct Answer: 2. (c)

Solution:

(c) Change in momentum,

$\begin{aligned} Δ p & = \int Fdt \\ = Area of F-t graph \\ = ar of Δ - ar of ◻ + ar of ◻ \\ = \frac{1}{2} \times 2 \times 6 - 3 \times 2 + 4 \times 3 = 12 N - s \end{aligned}$

According to Newton’s 2nd law. $F \propto \frac{d \vec{p}}{d t} \Rightarrow F$ $= K \frac{d \vec{p}}{d t}$ or, $F = \frac{d \vec{p}}{d t} [∵ k = 1$ in S.I & CGS system $]$ And $d \vec{p} = \vec{F} d t$

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Laws of Motion - Result Question 2

2. The force ’ F ’ acting on a particle of mass ’ m ’ is indicated by the force-time graph shown below. The change in momentum of the particle over the time interval from zero to 8s is :

Answer:

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2. The force ’ $F$ ’ acting on a particle of mass ’ $m$ ’ is indicated by the force-time graph shown below. The change in momentum of the particle over the time interval from zero to $8 s$ is :